notes on the octonions - geometry &...

Proceedings of 23rd GokovaGeometry-Topology Conferencepp. 1 – 85

Notes on the octonions

Dietmar A. Salamon* and Thomas Walpuski

Abstract. This is an expository paper. Its purpose is to explain the linear algebra

that underlies Donaldson–Thomas theory and the geometry of Riemannian manifolds

with holonomy in G2 and Spin(7).

1. Introduction

In these notes we give an exposition of the structures in linear algebra that under-lie Donaldson–Thomas theory [DT98, DS11] and calibrated geometry [HL82, Joy00]. Noclaim is made to originality. All the results and ideas described here (except perhaps The-orem 7.8) can be found in the existing literature, notably in the beautiful paper [HL82]by Harvey and Lawson. Perhaps these notes might be a useful introduction for studentswho wish to enter the subject.

Our emphasis is on characterizing the relevant algebraic structures—such as crossproducts, triple cross products, associator and coassociator brackets, associative, coas-socitative, and Cayley calibrations and subspaces—by their intrinsic properties ratherthan by the existence of isomorphisms to the standard structures on the octonions andthe imaginary octonions, although both descriptions are of course equivalent.

Section 2 deals with cross products and their associative calibrations. It contains aproof that they exist only in dimensions 0, 1, 3, and 7. In Section 3 we discuss nonde-generate 3–forms on 7–dimensional vector spaces (associative calibrations) and explainhow they give rise to unique compatible inner products. Additional structures such asassociative and coassociative subspaces and the associator and coassociator brackets arediscussed in Section 4. These structures are relevant for understanding G2–structures on7–manifolds and the Chern–Simons functional in Donaldson–Thomas theory.

The corresponding Floer theory has as its counterpart in linear algebra the productwith the real line. This leads to the structure of a normed algebra which only exists in di-mensions 1, 2, 4, and 8, corresponding to the reals, the complex numbers, the quaternions,and the octonions. These structures are discussed in Section 5. Going from Floer theoryto an intrinsic theory for Donaldson-type invariants of 8–dimensional Spin(7)–manifoldscorresponds to dropping the space-time splitting. The algebraic counterpart of this re-duction is to eliminate the choice of the unit (as well as the product). What is left of

Received by the editors 2017-02-01.

* partially supported by the Swiss National Science Foundation

1

SALAMON and WALPUSKI

the algebraic structures is the triple cross product and its Cayley calibration—a suitable4–form on an 8–dimensional Hilbert space. These structures are discussed in Section 6.Section 7 characterizes those 4–forms on 8–dimensional vector spaces (the Cayley-forms)that give rise to (unique) compatible inner products and hence to triple cross products.The relevant structure groups G2 (in dimension 7) and Spin(7) (in dimension 8) are dis-cussed in Section 8 and Section 9 with a particular emphasis on the splitting of the spaceof alternating multi-linear forms into irreducible representations. In Section 10 we exam-ine spin structures in dimensions 7 and 8. Section 11 relates SU(3) and SU(4) structuresto cross products and triple cross products and Section 12 gives a brief introduction tothe basic setting of Donaldson–Thomas theory.

Here is a brief overview of some of the literature about the groups G2 and Spin(7). Theconcept of a calibration was introduced in the article of Harvey–Lawson [HL82] which alsocontains definitions of G2 and Spin(7) in terms of the octonions. Humphreys [Hum78,Section 19.3] constructs (the Lie algebra of) G2 from the Dynkin diagram and proves thatthis coincides with the definition in terms of the octonions. The characterization of G2

and Spin(7) as the stabilisers of certain 3– and 4–forms is due to Bonan [Bon66]. Theconnection between calibrations and spinors is discussed in Harvey’s book [Har90] as wellas in the article of Dadok–Harvey [DH93].

Harvey–Lawson also introduced the (multiple) cross products and the associator andcoassociator brackets. The concept of a multiple cross product goes back to Eckmann[Eck43]. Building on this work, Whitehead [Whi62] classified those completely; see alsoBrown–Gray [BG67]. To our best knowledge, the splitting of the exterior algebra intoirreducible G2–representations is due to Fernandez–Gray [FG82, Section 3], who alsoemphasize the relation between G2 and the cross product in dimension seven. This aswell as the analogous result for Spin(7) can also be found in Bryant [Bry87, Section 2].

Among many others, the more recent articles by Bryant [Bry06], Karigiannis [Kar08,Kar09, Kar10] and Munoz [Mun14, Section 2] contain useful summaries of the linearalgebra related to G2 and Spin(7).

2. Cross products

We assume throughout that V is a finite dimensional real Hilbert space.

Definition 2.1. A skew-symmetric bilinear map

V × V → V : (u, v) 7→ u× v (2.2)

is called a cross product if it satisfies

〈u× v, u〉 = 〈u× v, v〉 = 0, and (2.3)

|u× v|2 = |u|2|v|2 − 〈u, v〉2 (2.4)

for all u, v ∈ V .

A bilinear map (2.2) that satisfies (2.4) also satisfies u×u = 0 for all u ∈ V and, hence,is necessarily skew-symmetric.

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Theorem 2.5. V admits a cross product if and only if its dimension is either 0, 1, 3,or 7. In dimensions 0 and 1 the cross product vanishes, in dimension 3 it is unique upto sign and determined by an orientation of V , and in dimension 7 it is unique up toorthogonal isomorphism.

Proof. See page 7.

The proof of Theorem 2.5 is based on the next five lemmas.

Lemma 2.6. Let (2.2) be a skew-symmetric bilinear map. Then the following are equiv-alent:

(i) Equation (2.3) holds for all u, v ∈ V .(ii) For all u, v, w ∈ V we have

〈u× v, w〉 = 〈u, v × w〉. (2.7)

(iii) The map φ : V 3 → R, defined by

φ(u, v, w) := 〈u× v, w〉, (2.8)

is an alternating 3–form (called the associative calibration of (V,×)).

Proof. Let (2.2) be a skew-symmetric bilinear map. Assume that it satisfies (2.3). Then,for all u, v, w ∈ V , we have

0 = 〈v × (u+ w), u+ w〉= 〈v × w, u〉+ 〈v × u,w〉= 〈u, v × w〉 − 〈u× v, w〉.

This proves (2.7).Now assume (2.7) and let φ be defined by (2.8). Then, by skew-symmetry, we have

φ(u, v, w) + φ(v, u, w) = 0 for all u, v, w and, by (2.7), we have φ(u, v, w) = φ(v, w, u) forall u, v, w. Hence, φ is an alternating 3–form. Thus we have proved that (i) implies (ii)implies (iii).

That (iii) implies (i) is obvious. This proves Lemma 2.6.

Lemma 2.9. Let (2.2) be a skew-symmetric bilinear map that satisfies (2.3). Then thefollowing are equivalent:

(i) The bilinear map (2.2) satisfies (2.4).(ii) If u and w are orthonormal, then |u× w| = 1.

(iii) If |u| = 1 and w is orthogonal to u, then u× (u× w) = −w.(iv) For all u,w ∈ V we have

u× (u× w) = 〈u,w〉u− |u|2w. (2.10)

(v) For all u, v, w ∈ V we have

u× (v × w) + v × (u× w) = 〈u,w〉v + 〈v, w〉u− 2〈u, v〉w. (2.11)

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Proof. That (i) implies (ii) is obvious.We prove that (ii) implies (iii). Fix a vector u ∈ V with |u| = 1 and define the linear

map A : V → V by Aw := u× w. Then, by skew-symmetry and (2.7), A is skew-adjointand, by (2.3), it preserves the subspace W := u⊥. Hence, the restriction of A2 to W is

self-adjoint and, by (ii), it satisfies 〈w,A2w〉 = −|u×w|2 = −|w|2 for w ∈W . Hence, therestriction of A2 to W is equal to minus the identity. This proves that (ii) implies (iii).

We prove that (iii) implies (iv). Fix a vector u ∈ V and define A : V → V by

Aw := u× w as above. By (iii) we have A2w = −|u|2w whenever w is orthogonal to u.Since A2u = 0, this implies (iv).

Assertion (v) follows from (iv) by replacing u with u+v. To prove that (v) implies (i),set w = v in (2.11) and take the inner product with u. Then

|u× v|2 = 〈u, u× (v × v) + v × (u× v)〉 = |u|2|v|2 − 〈u, v〉2.

Here the first equality follows from (2.7) and the second from (2.11) with w = v. Thisproves Lemma 2.9.

Lemma 2.12. Assume dimV = 3.

(i) A cross product on V determines a unique orientation such that u, v, u × v form apositive basis for every pair of linearly independent vectors u, v ∈ V .

(ii) If (2.2) is a cross product on V , then the 3–form φ given by (2.8) is the volume formassociated to the inner product and the orientation in (i).

(iii) If (2.2) is a cross product on V , then

(u× v)× w = 〈u,w〉v − 〈v, w〉u (2.13)

for all u, v, w ∈ V .(iv) Fix an orientation on V and denote by φ ∈ Λ3V ∗ the associated volume form.

Then (2.8) determines a cross product on V .

Proof. Assertion (i) follows from the fact that the space of pairs of linearly independentvectors in V is connected (whenever dimV 6= 2). Assertion (ii) follows from the fact that,if u, v are orthonormal, then u, v, u× v form a positive orthonormal basis and

φ(u, v, u× v) = |u× v|2 = 1.

We prove (iii). If u and v are linearly dependent, then both sides of (2.13) vanish. Hencewe may assume that u and v are linearly independent or, equivalently, that u × v 6= 0.Since (u × v) × w is orthogonal to u × v, by equation (2.7), and V has dimension 3, itfollows that (u × v) × w must be a linear combination of u and v. The formula (2.13)follows by taking the inner products with u and v, and using Lemma 2.9 (v).

We prove (iv). Assume that the bilinear map (2.2) is defined by (2.8), where φ is thevolume form associated to an orientation of V . Then skew-symmetry and (2.3) followfrom the fact that φ is a 3–form (see Lemma 2.6). If u, v are linearly independent, then

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by (2.8) we have u × v 6= 0 and φ(u, v, u× v) = |u× v|2 > 0. If u, v are orthonormal, itfollows that u, v, u× v is a positive orthogonal basis and so

φ(u, v, u× v) = |u× v|.Combining these two identities we obtain |u× v| = 1 when u, v are orthonormal. Hence,(2.4) follows from Lemma 2.9. This proves Lemma 2.12.

Example 2.14. On R3 the cross product associated to the standard inner product andthe standard orientation is given by the familiar formula

u× v =

u2v3 − u3v2

u3v1 − u1v3

u1v2 − u2v1

.

Example 2.15. The standard structure on R7 can be obtained from a basis of the formi, j,k, e, ei, ej, ek, where i, j,k, e are anti-commuting generators with square minus oneand ij = k. Then the cross product is given by

u× v :=

u2v3 − u3v2 − u4v5 + u5v4 − u6v7 + u7v6

u3v1 − u1v3 − u4v6 + u6v4 − u7v5 + u5v7

u1v2 − u2v1 − u4v7 + u7v4 − u5v6 + u6v5

u1v5 − u5v1 + u2v6 − u6v2 + u3v7 − u7v3

−u1v4 + u4v1 − u2v7 + u7v2 + u3v6 − u6v3

u1v7 − u7v1 − u2v4 + u4v2 − u3v5 + u5v3

−u1v6 + u6v1 + u2v5 − u5v2 − u3v4 + u4v3

. (2.16)

Witheijk := dxi ∧ dxj ∧ dxk

the associated 3–form (2.8) is given by

φ0 = e123 − e145 − e167 − e246 − e275 − e347 − e356. (2.17)

The product (2.16) is skew-symmetric and (2.7) follows from the fact that the matrixA(u) defined by

A(u)v := u× vis skew symmetric for all u, namely,

A(u) :=

0 −u3 u2 u5 −u4 u7 −u6

u3 0 −u1 u6 −u7 −u4 u5

−u2 u1 0 u7 u6 −u5 −u4

−u5 −u6 −u7 0 u1 u2 u3

u4 u7 −u6 −u1 0 u3 −u2

−u7 u4 u5 −u2 −u3 0 u1

u6 −u5 u4 −u3 u2 −u1 0

.

We leave it to the reader to verify (2.4) (or equivalently |u× v| = 1 whenever u and v areorthonormal).

See also Remark 3.6 below.

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Lemma 2.18. Let V be a be a real Hilbert space and (2.2) be a cross product on V . Letφ ∈ Λ3V ∗ be given by (2.8). Then the following holds:

(i) Let u ∈ V be a unit vector and Wu := u⊥. Define ωu : Wu ×Wu → R andJu : Wu →Wu by

ωu(v, w) := 〈u, v × w〉, Juv := u× v

for v, w ∈ Wu. Then ωu is a symplectic form on Wu, Ju is a complex structurecompatible with ωu, and the associated inner product is the one inherited from V .In particular, the dimension of V is odd.

(ii) Suppose dimV = 2n+ 1 ≥ 3. Then there is a unique orientation of V such that theassociated volume form vol ∈ Λ2n+1V ∗ satisfies

(ι(u)φ)n−1 ∧ φ = n!|u|n−1

vol (2.19)

for every u ∈ V . In particular, n is odd.

Proof. We prove (i). By Lemma 2.6 the bilinear form ωu is skew symmetric and, byLemma 2.9, we have Ju Ju = −1. Moreover,

ωu(v, Juw) = 〈u× v, u× w〉 = −〈v, u× (u× w)〉 = 〈v, w〉

for all v, w ∈ V . Here the first equation follows from the definition of ωu and Ju, thesecond follows from (2.7), and the last from Lemma 2.9. Thus the dimension of Wu iseven and so the dimension of V is odd.

We prove (ii). The set of all bases (u, v1, . . . , v2n) ∈ V 2n+1, where u has norm one andv1, . . . , v2n is a symplectic basis of Wu, is connected. Hence, there is a unique orientationof V with respect to which every such basis is positive. Let vol ∈ Λ2n+1V ∗ be theassociated volume form. To prove equation (2.19) assume first that |u| = 1 and choosean orthonormal symplectic basis v1, . . . , v2n of Wu. (For example pick an orthonormalbasis v1, v3, . . . , v2n−1 of a Lagrangian subspace of Wu and define v2k := Juv2k−1 fork = 1, . . . , n.) Now evaluate both sides of the equation on the tuple (u, v1, . . . , v2n). Thenwe obtain n! on both sides. This proves (2.19) whenever u has norm one. The generalcase follows by scaling. It follows from (2.19) that n is odd since otherwise the left handside changes sign when we replace u by −u. This proves Lemma 2.18.

Lemma 2.20. Let n > 1 be an odd integer and V be an oriented real Hilbert space ofdimension 2n + 1 with volume form vol ∈ Λ2n+1V ∗. Let φ ∈ Λ3V ∗ be a 3–form anddenote its isotropy group by

G := g ∈ Aut(V ) : g∗φ = φ .

If φ satisfies (2.19), then G ⊂ SO(V ).

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Proof. Let g ∈ G and u ∈ V . Then it follows from (2.19) that

|gu|n−1g∗vol =

1

n!g∗(

(ι(gu)φ)n−1 ∧ φ

)=

1

n!

((g∗ι(gu)φ)

n−1 ∧ g∗φ)

=1

n!(ι(u)g∗φ)

n−1 ∧ g∗φ

=1

n!(ι(u)φ)

n−1 ∧ φ

= |u|n−1vol.

Hence, there is a constant c > 0 such that

g∗vol = c−1vol, |gu|n−1= c|u|n−1

for every u ∈ V . Since n > 1, this gives |gu| = c1

n−1 |u| for u ∈ V and hence

g∗vol = c2n+1n−1 vol = c

3nn−1 g∗vol.

Thus c = 1 and this proves Lemma 2.20.

Proof of Theorem 2.5. Assume dimV > 1, let (2.2) be a cross product on V , and defineφ : V ×V ×V → R by (2.8). By Lemma 2.6, we have φ ∈ Λ3V ∗. By Lemma 2.18 (i), thedimension of V is odd. By Lemma 2.20, we have dimV = 4n+ 3 for some integer n ≥ 0.In particular dimV 6= 5.

We prove that dimV ≤ 7. Define A : V → End(V ) by A(u)v := u× v. Then it followsfrom Lemma 2.9 that

A(u)u = 0, A(u)2 = uu∗ − |u|21.

Define γ : V → End(R⊕ V ) by

γ(u) :=

(0 −u∗u A(u)

), (2.21)

where u∗ : V → R denotes the linear functional v 7→ 〈u, v〉. Then

γ(u)∗ + γ(u) = 0, γ(u)∗γ(u) = |u|21 (2.22)

for every u ∈ V . Here the first equation follows from the fact that A(u) is skew-adjoint forevery u and the last equation follows by direct calculation. This implies that γ extendsto a linear map from the Clifford algebra C`(V ) to End(R ⊕ V ). The restriction of thisextension to the Clifford algebra of any even dimensional subspace of V is injective (see,e.g. [Sal99, Proposition 4.13]). Hence, 22n ≤ (2n + 2)2. This implies n ≤ 3 and sodimV = 2n + 1 ≤ 7. Thus we have proved that the dimension of V is either 0, 1, 3, or7. That the cross product vanishes in dimension 0 and 1 is obvious. That it is uniquelydetermined by the orientation of V in dimension 3 follows from Lemma 2.12. The lastassertion of Theorem 2.5 is restated and proved in Theorem 3.2 below.

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Remark 2.23. Let V be a nonzero real Hilbert space that admits a 3–form φ whoseisotropy subgroup G is contained in SO(V ). Then

dim Aut(V )− dim Λ3V ∗ ≤ dim G ≤ dim SO(V ).

Hence, dimV ≥ 7 as otherwise dim SO(V ) < dim Aut(V )−dim Λ3V ∗. This gives anotherproof for the nonexistence of cross products in dimension 5.

3. Associative calibrations

Definition 3.1. Let V be a real vector space. A 3–form φ ∈ Λ3V ∗ is called nondegen-erate if, for every pair of linearly independent vectors u, v ∈ V , there is a vector w ∈ Vsuch that φ(u, v, w) 6= 0. An inner product on V is called compatible with φ if themap (2.2) defined by (2.8) is a cross product.

Theorem 3.2. Let V be a 7–dimensional real vector space and φ, φ′ ∈ Λ3V ∗. Then thefollowing holds:

(i) φ is nondegenerate if and only if it admits a compatible inner product.(ii) The inner product in (i), if it exists, is uniquely determined by φ.

(iii) If φ and φ′ are nondegenerate, the vectors u, v, w are orthonormal for φ and sat-isfy φ(u, v, w) = 0, and the vectors u′, v′, w′ are orthonormal for φ′ and satisfyφ′(u′, v′, w′) = 0, then there exists a g ∈ Aut(V ) such that g(u) = u′, g(v) = v′,g(w) = w′, and g∗φ′ = φ.

Proof. See pages 11 and 12.

Remark 3.3. If dimV = 3, then φ ∈ Λ3V ∗ is nondegenerate if and only if it is nonzero.If φ 6= 0, then, by Lemma 2.12, an inner product on V is compatible with φ if and onlyif φ is the associated volume form with respect to some orientation, i.e., φ(u, v, w) = ±1for every orthonormal basis u, v, w of V . Thus assertion (i) of Theorem 3.2 continues tohold in dimension three.

However, assertion (ii) is specific to dimension seven.

Lemma 3.4. Let V be a 7–dimensional real Hilbert space and φ ∈ Λ3V ∗. Then thefollowing are equivalent:

(i) φ is compatible with the inner product.(ii) There is an orientation on V such that the associated volume form vol ∈ Λ7V ∗

satisfies

ι(u)φ ∧ ι(v)φ ∧ φ = 6〈u, v〉vol (3.5)


Each of these conditions implies that φ is nondegenerate. Moreover, the orientation in (ii),if it exists, is uniquely determined by φ.

Remark 3.6. It is convenient to use equation (3.5) to verify that the bilinear map inExample 2.15 satisfies (2.4). In fact, it suffices to check (3.5) for every pair of standard

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basis vectors. Care must be taken. There are examples of 3–forms φ on V = R7 for whichthe quadratic form

V × V → Λ7V ∗ : (u, v) 7→ ι(u)φ ∧ ι(v)φ ∧ φhas signature (3, 4). One such example can be obtained from the 3–form φ0 in Exam-ple 2.15 by changing the minus signs to plus.

Proof of Lemma 3.4. If (i) holds, then, by Lemma 2.18 (ii), there is a unique orientationon V such that the associated volume form satisfies

ι(u)φ ∧ ι(u)φ ∧ φ = 6|u|2vol

for every u ∈ V . Applying this identity to u + v and u − v and taking the difference weobtain (3.5). Moreover, if u, v ∈ V are linearly independent, then

φ(u, v, u× v) = |u× v|2 = |u|2|v|2 − 〈u, v〉2 6= 0.

Hence, φ is nondegenerate. This shows that (i) implies (ii) and nondegeneracy.Conversely, assume (ii). We prove that φ is nondegenerate. Let u, v ∈ V be linearly

independent. Then u 6= 0 and, hence, by (3.5), the 7–form

σ := ι(u)φ ∧ ι(u)φ ∧ φ = 6|u|2vol ∈ Λ7V ∗

is nonzero. Choose a basis v1, . . . , v7 of V with v1 = u and v2 = v. Evaluating σ on thisbasis we obtain that one of the terms φ(u, v, vj) with j ≥ 3 must be nonzero. Hence, φ isnondegenerate as claimed.

Now define the bilinear map V × V → V : (u, v) 7→ u× v by (2.8). This map is skew-symmetric and, by Lemma 2.6, it satisfies (2.3). We must prove that it also satisfies (2.4).By Lemma 2.9, it suffices to show

|u| = 1, 〈u, v〉 = 0 =⇒ |u× v| = |v|. (3.7)

We prove this in five steps. Throughout we fix a unit vector u ∈ V .

Step 1. Define the linear map A : V → V by Av := u× v. Then A is skew-adjoint andits kernel is spanned by u.

That A is skew-adjoint follows from the identity 〈Av,w〉 = φ(u, v, w). That its kernelis spanned by u follows from the fact that φ is nondegenerate.

Step 2. Let A be as in Step 1. Then there are positive constants λ1, λ2, λ3 and anorthonormal basis v1, w1, v2, w2, v3, w3 of u⊥ such that Avj = λjwj and Awj = −λjvj forj = 1, 2, 3.

By Step 1, there is a constant λ > 0 and a vector v ∈ u⊥ such that

A2v = −λ2v, |v| = 1.

Denote w := λ−1Av. Then Av = λw, Aw = −λv, w is orthogonal to v, and

|w|2 = λ−2〈Av,Av〉 = −λ−2〈v,A2v〉 = |v|2 = 1.

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Moreover, the orthogonal complement of u, v, w is invariant under A. Hence, Step 2follows by induction.

Step 3. Let λi be as in Step 2. Then λ1λ2λ3 = 1.

Let A be as in Step 1, denote W := u⊥, and define ω : W ×W → R by

ω(v, w) := 〈Av,w〉 = φ(u, v, w)

for v, w ∈W . Then, by Step 1, ω ∈ Λ2W ∗ is a symplectic form. Moreover,

ω(vi, wi) = 〈Avi, wi〉 = λi

for i = 1, 2, 3 while ω(vi, wj) = 0 for i 6= j and

ω(vi, vj) = ω(wi, wj) = 0

for all i and j. Hence,

λ1λ2λ3 =1

6ω3(v1, w1, v2, w2, v3, w3)

= vol(u, v1, w1, v2, w2, v3, w3).

Here the first equation follows from Step 2 and the definition of ω and the second equationfollows from (3.5) with u = v and |u| = 1. Since the vectors u, v1, w1, v2, w2, v3, w3 forman orthonormal basis of V , the last expression must be plus or minus one. Since it ispositive, Step 3 follows.

Step 4. Define

G := g ∈ Aut(V ) : g∗φ = φ , H := g ∈ G : gu = u . (3.8)

Then dim G ≥ 14 and dim H ≥ 8

Since dim Aut(V ) = 49 and dim Λ3V ∗ = 35, the isotropy subgroup G of φ has dimen-sion at least 14. Moreover, by Lemma 2.20, G acts on the sphere S := v ∈ V : |v| = 1which has dimension 6. Thus the isotropy subgroup H of u under this action has dimensiondim H ≥ dim G− dimS ≥ 14− 6 = 8. This proves Step 4.

Step 5. Let λi be as in Step 2. Then λ1 = λ2 = λ3 = 1.

By definition of A in Step 1 and H in Step 4, we have 〈Agv, gw〉 = 〈Av,w〉 for all g ∈ Hand all v, w ∈ V . Moreover, H ⊂ SO(V ), by Lemma 2.20. Hence,

g ∈ H =⇒ gA = Ag. (3.9)

Now suppose that the eigenvalues λ1, λ2, λ3 are not all equal. Without loss of generality,we may assume λ1 /∈ λ2, λ3. Then, by (3.9), the subspaces W1 := spanv1, w1 andW23 := spanv2, w2, v3, w3 are preserved by each element g ∈ H. Thus

H ⊂ O(W1)×O(W23).

Since dim O(W1) = 1 and dim O(W23) = 6, this implies dim H ≤ 7 in contradiction toStep 4. Thus we have proved that λ1 = λ2 = λ3 and, by Step 3, this implies λj = 1 forevery j. This proves Step 5.

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By Step 2 and Step 5 we have A2v = −v for every v ∈ u⊥. Hence, by Step 1,|Av|2 = −〈v,A2v〉 = |v|2 for every v ∈ u⊥. By definition of A, this proves (3.7) andLemma 3.4.

Proof of Theorem 3.2 (i) and (ii). The “if” part of (i) is the last assertion made inLemma 3.4. To prove (ii) and the “only if” part of (i) we assume that φ is nondegenerate.Then, for every nonzero vector u ∈ V , the restriction of the 2–form ι(u)φ ∈ Λ2V ∗ to u⊥

is a symplectic form. Namely, if v ∈ u⊥ is nonzero, then u, v are linearly independentand hence there is a vector w ∈ V such that φ(u, v, w) 6= 0; the vector w can be chosenorthogonal to u.

This implies that the restriction of the 6–form (ι(u)φ)3 ∈ Λ6V ∗ to u⊥ is nonzero for

every nonzero vector u ∈ V . Hence, the 7–form ι(u)φ ∧ ι(u)φ ∧ φ ∈ Λ7V ∗ is nonzero forevery nonzero vector u ∈ V . Since V \ 0 is connected, there is a unique orientation ofV such that ι(u)φ ∧ ι(u)φ ∧ φ is a positive volume form on V for every u ∈ V \ 0. Fixa volume form σ ∈ Λ7V ∗ compatible with this orientation. Then the bilinear form

V × V → R : (u, v) 7→ ι(u)φ ∧ ι(v)φ ∧ φσ

=: g(u, v)

is an inner product. Define µ > 0 by σ = µvolg. Replacing σ by σ := λ2σ we get

g = λ−2g, volg = λ−7volg.

Thus

σ = λ2σ = λ2µvolg = λ9µvolg.

With λ := (6/µ)1/9 we get σ = 6volg.Thus we have proved that there is a unique orientation and inner product on V such

that φ satisfies (3.5). Hence the assertion follows from Lemma 3.4. This proves parts (i)and (ii) of Theorem 3.2.

Remark 3.10. Let V,W be n-dimensional real vector spaces. Then the determinant ofa linear map A : V → W is an element detA ∈ ΛnV ∗ ⊗ ΛnW . In particular, if V isequipped with an orientation and an inner product g ∈ S2V ∗, and ig : V → V ∗ denotesthe isomorphism defined by igv := g(v, ·), then det ig ∈ (ΛnV ∗)2 and the volume formvolg associated to g is

volg =√

det ig.

Here the orientation is needed to determine the sign of the square root.If V is 7–dimensional and φ ∈ Λ3V ∗ is nondegenerate, then the formula

G(u, v) :=1

6i(u)φ ∧ i(v)φ ∧ φ for u, v ∈ V

defines a symmetric bilinear form G : V × V → Λ7V ∗ and iG : V → V ∗ ⊗ Λ7V ∗ is anisomorphism (see second paragraph in the proof of Lemma 3.4). The determinant of iG

11


is an element of (Λ7V ∗)9 and (det iG)1/9 can be defined without an orientation on V . Ifan inner product g and an orientation on V are such that (3.5) holds, then

volg = (det(iG))1/9 and g =G

volg.

Conversely, with this choice of inner product and orientation, (3.5) holds. This observationis due to Hitchin [Hit01, Section 8.3].

Lemma 3.11. Let V be a 7–dimensional real Hilbert space equipped with a cross productV × V → V : (u, v)→ u× v. If u and v are orthonormal and w := u× v, then v×w = uand w × u = v.

Proof. This follows immediately from equation (2.11) in Lemma 2.9.

Proof of Theorem 3.2 (iii). Let φ0 : R7 ×R7 ×R7 → R be the 3–form in Example 2.15and let φ ∈ Λ3V ∗ be a nondegenerate 3–form. Let V be equipped with the compatibleinner product of Theorem 3.2 and denote by V × V → V : (u, v) 7→ u× v the associatedcross product. Let e1, e2 ∈ V be orthonormal and define

e3 := e1 × e2.

Let e4 ∈ V be any unit vector orthogonal to e1, e2, e3 and define

e5 := −e1 × e4.

Then e5 has norm one and is orthogonal to e1, e2, e3, e4. For e1 and e4 this follows fromthe definition and (2.7). For e3 we observe

〈e3, e5〉 = −〈e1 × e2, e1 × e4〉 = 〈e2, e1 × (e1 × e4)〉 = −〈e2, e4〉 = 0.

Here the last but one equation follows from Lemma 2.9. For e2 the argument is similar;since e2 = e3 × e1, by Lemma 3.11, and 〈e3, e4〉 = 0, we obtain 〈e2, e5〉 = 0. Now let e6

be a unit vector orthogonal to e1, . . . , e5 and define

e7 := −e1 × e6.

As before we have that e7 has norm one and is orthogonal to e1, . . . , e6. Thus the vectorse1, . . . , e7 form an orthonormal basis of V and it follows from Lemma 3.11 that theysatisfy the same relations as the standard basis of R7 in Example 2.15. Hence, the map

R7 g−→ V : x = (x1, . . . , x7) 7→7∑i=1

xiei

is a Hilbert space isometry and it satisfies g∗φ = φ0. This proves Theorem 3.2 (and thelast assertion of Theorem 2.5).

12


4. The associator and coassociator brackets

We assume throughout that V is a 7–dimensional real Hilbert space, that φ ∈ Λ3V ∗ isa nondegenerate 3–form compatible with the inner product, and (2.2) is the cross productgiven by (2.8). It follows from (2.11) that the expression (u × v) × w is alternating onany triple of pairwise orthogonal vectors u, v, w ∈ V . Hence, it extends uniquely to analternating 3–form V 3 → V : (u, v, w) 7→ [u, v, w] called the associator bracket. Anexplicit formula for this 3–form is

[u, v, w] := (u× v)× w + 〈v, w〉u− 〈u,w〉v. (4.1)

The associator bracket can also be expressed in the form

[u, v, w] =1

3

((u× v)× w + (v × w)× u+ (w × u)× v

). (4.2)

Remark 4.3. If V is any Hilbert space with a skew-symmetric bilinear form (2.2), then theassociator bracket (4.1) is alternating if and only if (2.11) holds. Indeed, skew-symmetryof the associator bracket in the first two arguments is obvious, and the identity

[u, v, w] + [u,w, v] = w × (v × u) + v × (w × u)

− 〈u,w〉v − 〈u, v〉w + 2〈v, w〉u

shows that skew-symmetry in the last two arguments is equivalent to (2.11). ByLemma 2.12, the associator bracket vanishes in dimension three.

The square of the volume of the 3–dimensional parallelepiped spanned by u, v, w ∈ Vwill be denoted by

|u ∧ v ∧ w|2 := det

|u|2 〈u, v〉〈u,w〉〈v, u〉 |v|2 〈v, w〉〈w, u〉〈w, v〉 |w|2

.

Lemma 4.4. For all u, v, w ∈ V we have

φ(u, v, w)2 + |[u, v, w]|2 = |u ∧ v ∧ w|2. (4.5)

Proof. If w is orthogonal to u and v, then we have

|[u, v, w]|2 = |(u× v)× w|2

= |u× v|2|w|2 − 〈u, v × w〉2

= |u ∧ v ∧ w|2 − φ(u, v, w)2.

Here the first equation follows from the definition of the associator bracket and orthog-onality, the second equation follows from (2.4), and the last equation follows from (2.4)and orthogonality, as well as (2.8). The general case can be reduced to the orthogonalcase by Gram–Schmidt.

13


Definition 4.6. A 3–dimensional subspace Λ ⊂ V is called associative the associatorbracket vanishes on Λ, i.e.,

[u, v, w] = 0 for all u, v, w ∈ Λ.

Lemma 4.7. Let Λ ⊂ V be a 3–dimensional linear subspace. Then the following areequivalent:

(i) Λ is associative.(ii) If u, v, w is an orthonormal basis of Λ, then φ(u, v, w) = ±1.

(iii) If u, v ∈ Λ, then u× v ∈ Λ.(iv) If u ∈ Λ⊥ and v ∈ Λ, then u× v ∈ Λ⊥.(v) If u, v ∈ Λ⊥, then u× v ∈ Λ.

Moreover, if u, v ∈ V are linearly independent, then the subspace spanned by the vectorsu, v, u× v is associative.

Proof. That (i) is equivalent to (ii) follows from Lemma 4.4.We prove that (i) is equivalent to (iii). That the associator bracket vanishes on a 3–

dimensional subspace that is invariant under the cross product follows fromLemma 2.12 (iii). Conversely suppose that the associator bracket vanishes on Λ. Letu, v ∈ Λ be linearly independent and let w ∈ Λ be a nonzero vector orthogonal to u andv. Then, by Lemma 4.4, we have

〈u× v, w〉2 = φ(u, v, w)2 = |u ∧ v ∧ w|2 = |u× v|2|w|2

and hence u× v is a real multiple of w. Thus u× v ∈ Λ.We prove that (iii) is equivalent to (iv). First assume (iii) and let u ∈ Λ, v ∈ Λ⊥.

Then, by (iii), we have w× u ∈ Λ for every w ∈ Λ. Hence, 〈w, u× v〉 = 〈w× u, v〉 = 0 forevery w ∈ Λ and so u× v ∈ Λ⊥. Conversely assume (iv) and let u, v ∈ Λ. Then, by (iii),we have w × u ∈ Λ⊥ for every w ∈ Λ⊥. Hence,

〈w, u× v〉 = 〈w × u, v〉 = 0

for every w ∈ Λ⊥. This implies u × v ∈ Λ. Thus we have proved that (iii) is equivalentto (iv).

We prove that (iv) is equivalent to (v). Fix a unit vector u ∈ Λ⊥ and define theendomorphism J : u⊥ → u⊥ by

Jv := u× v.By Lemma 2.9 this is an isomorphism with inverse −J . Condition (iv) asserts that Jmaps Λ to Λ⊥∩u⊥ while condition (v) asserts that J maps Λ⊥∩u⊥ to Λ. Since both are3–dimensional subspaces of u⊥, these two assertions are equivalent. This proves that (iv)is equivalent to (v).

If u and v are linearly independent, then u×v 6= 0, by (2.4), and u× v is orthogonal tou and v, by (2.3). Hence, the subspace Λ spanned by u, v, u× v is 3–dimensional. Thatit is invariant under the cross product follows from assertion (iv) in Lemma 2.9. Hence,Λ is associative, and this proves Lemma 4.7.

14


Lemma 4.8. The map ψ : V 4 → R defined by

ψ(u, v, w, x) := 〈[u, v, w], x〉

=1

3

(φ(u× v, w, x) + φ(v × w, u, x) + φ(w × u, v, x)

) (4.9)

is an alternating 4–form (the coassociative calibration of (V, φ)). Moreover, it isgiven by ψ = ∗φ, where ∗ : ΛkV ∗ → Λ7−kV ∗ denotes the Hodge ∗–operator associated tothe inner product and the orientation in Lemma 3.4.

Proof. See page 15.

Remark 4.10. By Lemma 4.7 and Lemma 4.8 the associator bracket [u, v, w] is orthogonalto the vectors u, v, w, v×w,w×u, u×v. Second, these six vectors are linearly independentif only if [u, v, w] 6= 0. (Make them pairwise orthogonal by adding to v a real multipleof u and to w a linear combination of u, v, u× v. Then their span and [u, v, w] remainunchanged.) Third, if [u, v, w] 6= 0 then the vectors u, v, w, v × w,w × u, u × v, [u, v, w]form a positive basis of V .

Remark 4.11. The standard associative calibration on R7 is

φ0 = e123 − e145 − e167 − e246 + e257 − e347 − e356 (4.12)

(see Example 2.15). The corresponding coassociative calibration is

ψ0 = −e1247 − e1256 + e1346 − e1357 − e2345 − e2367 + e4567. (4.13)

Remark 4.14. Let V → V ∗ : u 7→ u∗ := 〈u, ·〉 be the isomorphism induced by the innerproduct. Then, for α ∈ ΛkV ∗ and u ∈ V , we have

∗ ι(u)α = (−1)k−1u∗ ∧ ∗α. (4.15)

This holds on any finite dimensional oriented Hilbert space.

Remark 4.16. Throughout we use the notation

(LAα)(v1, . . . , vk) := α(Av1, v2, . . . , vk) + · · ·+ α(v1, . . . , vk−1, Avk) (4.17)

for the infinitesimal action of A ∈ End(V ) on a k–form α ∈ ΛkV ∗. For u ∈ V denote byAu ∈ so(V ) the skew-adjoint endomorphism Auv := u× v. Then equation (4.9) can beexpressed in the form

LAuφ = 3ι(u)ψ. (4.18)

Since ψ = ∗φ, we have LAψ = ∗LAφ for all A ∈ so(V ). Hence, it follows from equa-tion (4.15) that

LAuψ = ∗(3ι(u)ψ) = −3u∗ ∧ φ. (4.19)

Proof of Lemma 4.8. It follows from Remark 4.3 that ψ is alternating in the first threearguments. To prove that ψ ∈ Λ4V ∗ we compute

ψ(u, v, w, x) = 〈(u× v)× w + 〈v, w〉u− 〈u,w〉v, x〉= 〈u× v, w × x〉+ 〈v, w〉〈u, x〉 − 〈u,w〉〈v, x〉.

(4.20)

15


Here the first equation follows from the definition of ψ in (4.9) and the definition of theassociator bracket in (4.1). Swapping x and w as well as u and v in (4.20) gives the sameexpression. Thus

ψ(u, v, w, x) = ψ(v, u, x, w) = −ψ(u, v, x, w).

This shows that ψ ∈ Λ4V ∗ as claimed. To prove the second assertion we observe thefollowing.

Claim. If u, v, w, x are orthonormal and u× v = w × x, then ψ(u, v, w, x) = 1.

This follows directly from the definition of ψ and of the associator bracket in (4.1)and (4.9). Now, by Theorem 3.2, we can restrict attention to the standard structureson R7. Thus φ = φ0 is given by (4.12) and this 3–form is compatible with the standardinner product on R7. We have the product rule ei × ej = ek whenever the term eijk orone of its cyclic permutations shows up in this sum, and the claim shows that we have asummand εeijk` in ψ = ψ0 whenever ei× ej = εek× e` with ε ∈ ±1. Hence, ψ0 is givenby (4.13). Term by term inspection shows that ψ0 = ∗φ0. This proves Lemma 4.8.

Lemma 4.21. For all u, v, w, x ∈ V we have

[u, v, w, x] := φ(u, v, w)x− φ(x, u, v)w + φ(w, x, u)v − φ(v, w, x)u

=1

3

(−[u, v, w]× x+ [x, u, v]× w − [w, x, u]× v + [v, w, x]× u

).

(4.22)

The resulting multi-linear map

V 4 → V : (u, v, w, x) 7→ [u, v, w, x]

is alternating and is called the coassociator bracket on V .

Proof. Define the alternating multi-linear map τ : V 4 → V by

τ(u, v, w, x) := 3(φ(u, v, w)x− φ(x, u, v)w + φ(w, x, u)v − φ(v, w, x)u

)+ [u, v, w]× x− [x, u, v]× w + [w, x, u]× v − [v, w, x]× u.

We must prove that τ vanishes. The proof has three steps.

Step 1. τ(u, v, w, x) is orthogonal to u, v, w, x for all u, v, w, x ∈ V .

It suffices to assume that u, v, w, x are pairwise orthogonal. Then we have

[u, v, w] = (u× v)× wand similarly for [x, v, w] etc. Hence,

〈τ(u, v, w, x), x〉 = 3|x|2φ(u, v, w)− 〈[u, v, x], w × x〉− 〈[w, u, x], v × x〉 − 〈[v, w, x], u× x〉

= 3|x|2φ(u, v, w)− 〈(u× v)× x,w × x〉− 〈(w × u)× x, v × x〉 − 〈(v × w)× x, u× x〉

= 0.

16


Here the last step uses the identity (2.7) and the fact that x× (u× x) = |x|2u wheneveru is orthogonal to x. Thus τ(u, v, w, x) is orthogonal to x. Since τ is alternating, thisproves Step 1.

Step 2. τ(u, v, u× v, x) = 0 for all u, v, x ∈ V .

It suffices to assume that u, v are orthonormal and that x is orthogonal to u, v, andw := u× v. Then v × w = u, w × u = v, φ(u, v, w) = 1, and

φ(x, v, w) = φ(x,w, u) = φ(x, u, v) = 0.

Moreover, [u, v, w] = 0 and

[x, v, w] = [v, w, x] = (v × w)× x = u× x, [x, v, w]× u = x,

and similarly [x,w, u]× v = [x, u, v]× w = x. This implies that τ(u, v, w, x) = 0.

Step 3. τ(u, v, w, x) = 0 for all u, v, w, x ∈ V .

By the alternating property we may assume that u and v are orthonormal. Using thealternating property again and Step 2 we may assume that w is a unit vector orthogonalto u, v, u × v and that x is a unit vector orthogonal to u, v, w and v × w,w × u, u × v.This implies that

φ(u, v, w) = φ(x, v, w) = φ(x,w, u) = φ(x, u, v) = 0.

Hence, the vectors x× u, x× v, x× w form a basis of the orthogonal complement of thespace spanned by u, v, w, x. Each of these vectors is orthogonal to τ(u, v, w, x) and henceτ(u, v, w, x) = 0 by Step 1. This proves Lemma 4.21.

The square of the volume of the 4–dimensional parallelepiped spanned by u, v, w, x ∈ Vwill be denoted by

|u ∧ v ∧ w ∧ x|2 := det

|u|2 〈u, v〉〈u,w〉〈u, x〉〈v, u〉 |v|2 〈v, w〉〈v, x〉〈w, u〉〈w, v〉 |w|2 〈w, x〉〈x, u〉〈x, v〉〈x,w〉 |x|2

.

Lemma 4.23. For all u, v, w, x ∈ V we have

ψ(u, v, w, x)2 + |[u, v, w, x]|2 = |u ∧ v ∧ w ∧ x|2. (4.24)

Proof. The proof has four steps.

Step 1. If u, v, w, x are orthogonal, then

ψ(u, v, w, x)2 = 〈u× v, w × x〉2,

|[u, v, w, x]|2 = 〈u× v, w〉2|x|2 + 〈u× v, x〉2|w|2

+ 〈u,w × x〉2|v|2 + 〈v, w × x〉2|u|2,

|u ∧ v ∧ w ∧ x|2 = |u|2|v|2|w|2|x|2.

17


The first equation follows from (4.1) and (4.9), using (2.7). The other two equationsfollow immediately from the definitions.

Step 2. Equation (4.24) holds when u, v, w, x are orthogonal and, in addition, w and xare orthogonal to u× v.

Since [u, v, w] 6= 0, it follows from the assumptions and Lemma 4.7 that w × x is alinear combination of the vectors u, v, u× v. Hence, the assertion follows from Step 1.

Step 3. Equation (4.24) holds when u, v, w, x are orthogonal

Suppose, in addition, that w and x are orthogonal to u × v and replace x byxλ := x+ λu× v for λ ∈ R. Then ψ(u, v, w, xλ) is independent of λ and

|[u, v, w, xλ]|2 = |[u, v, w, x]|2 + λ2|u|2|v|2|w|2|u× v|2.Hence, it follows from Step 2 that (4.24) holds when u, v, w, x are orthogonal and, inaddition, w is orthogonal to u × v. This condition can be achieved by rotating the pair(w, x). This proves Step 3.

Step 4. Equation (4.24) holds always.

The general case follows from the orthogonal case via Gram–Schmidt, because bothsides of equation (4.24) remain unchanged if we add to any of the four vectors a multipleof any of the other three. This proves the lemma.

Definition 4.25. A 4–dimensional subspace H ⊂ V is called coassociative if

[u, v, w, x] = 0 for all u, v, w, x ∈ H.

Lemma 4.26. Let H ⊂ V be a 4–dimensional linear subspace. Then the following areequivalent:

(i) H is coassociative.(ii) If u, v, w, x is an orthonormal basis of H, then ψ(u, v, w, x) = ±1.

(iii) For all u, v, w ∈ H we have φ(u, v, w) = 0.(iv) If u, v ∈ H, then u× v ∈ H⊥.(v) If u ∈ H and v ∈ H⊥, then u× v ∈ H.

(vi) If u, v ∈ H⊥, then u× v ∈ H⊥.(vii) The orthogonal complement H⊥ is associative.

Proof. That (i) is equivalent to (ii) follows from Lemma 4.23.We prove that (i) is equivalent to (iii). That (iii) implies (i) is obvious by definition of

the coassociator bracket in (4.22). Conversely, assume (i) and choose a basis u, v, w, x ofH. Then [u, v, w, x] = 0 and hence, by (4.22), we have

φ(u, v, w) = φ(x, v, w) = φ(x,w, u) = φ(x, u, v) = 0.

This implies (iii).We prove that (iii) is equivalent to (iv). If (iii) holds and u, v ∈ H, then

〈u× v, w〉 = φ(u, v, w) = 0

18


for every w ∈ H and hence u × v ∈ H⊥. Conversely, if (iv) holds and u, v ∈ H, thenu× v ∈ H⊥ and hence φ(u, v, w) = 〈u× v, w〉 = 0 for all w ∈ H.

Thus we have proved that (i), (ii), (iii), (iv) are equivalent. That assertions (iv), (v),(vi), (vii) are equivalent was proved in Lemma 4.7.

Remark 4.27. Let V be a 7–dimensional real Hilbert space equipped with a cross productand denote the associative and coassociative calibrations by φ and ψ. Let Λ ⊂ V bean associative subspace and define H := Λ⊥. Orient Λ and H by the volume formsvolΛ := φ|Λ and volH := ψ|H . A standard basis of the space Λ+H∗ of self-dual 2–formson H is a triple ω1, ω2, ω3 ∈ Λ+H∗ that satisfies the condition ωi ∧ ωj = 2δijvolH for alli and j. In this situation the map

Λ→ Λ+H∗ : u 7→ −ι(u)φ|H (4.28)

is an orientation preserving isomorphism that sends every orthonormal basis of Λ to astandard basis of Λ+H∗. (To see this, choose a standard basis of V as in Remark 4.11 withΛ = spane1, e2, e3.) Let πΛ : V → Λ and πH : V → H be the orthogonal projections.Let u1, u2, u3 be any orthonormal basis of Λ and define

αi := u∗i |Λ and ωi := −ι(ui)φ|Hfor i = 1, 2, 3. Then the associative calibration φ can be expressed in the form

φ = π∗ΛvolΛ − π∗Λα1 ∧ π∗Hω1 − π∗Λα2 ∧ π∗Hω2 − π∗Λα3 ∧ π∗Hω3. (4.29)

The next theorem characterizes a nondegenerate 3–form φ in terms of its coassociativecalibration ψ in Lemma 4.8.

Theorem 4.30. Let V be a 7–dimensional vector space over the reals, let φ, φ′ ∈ Λ3V ∗

be nondegenerate 3–forms, and let ψ,ψ′ ∈ Λ4V ∗ be their coassociative calibrations. Thenthe following are equivalent:

(i) φ′ = φ or φ′ = −φ.(ii) ψ′ = ψ.

Proof. That (i) implies (ii) follows from the definition of ψ in Lemma 4.8 and the factthat reversing the sign of φ also reverses the sign of the cross product and thus leaves ψunchanged (see equation (4.9)). To prove the converse assume that ψ′ = ψ and denote by〈·, ·〉′ the inner product determined by φ′, by ×′ the cross product determined by φ′, andby [·, ·, ·]′ the associator bracket determined by φ′. We prove in four steps that φ′ = ±φ.

Step 1. A 3–dimensional subspace Λ ⊂ V is associative for φ if and only if it is associativefor φ′.

Let Λ ⊂ V be a three-dimensional linear subspace. By Definition 4.6 it is associativefor φ if and only if [u, v, w] = 0 for all u, v, w ∈ Λ. By Lemma 4.8 this is equivalent to thecondition that the linear functional ψ(u, v, w, ·) on V vanishes for all u, v, w ∈ Λ. Sinceψ = ψ′, this proves Step 1.

19


Step 2. There is a linear functional α : V → R and a c ∈ R \ 0 such that

u×′ v = α(u)v − α(v)u+ cu× v


Fix two linearly independent vectors u, v ∈ V . Then the vectors u, v, u × v span aφ–associative subspace Λ ⊂ V by Lemma 4.7. The subspace Λ is also φ′–associative byStep 1. Hence, u×′ v ∈ Λ by Lemma 4.7 and so there exist real numbers α(u, v), β(u, v),γ(u, v) such that

u×′ v = α(u, v)v + β(u, v)u+ γ(u, v)u× v. (4.31)

Since u, v, u×′ v are linearly independent, it follows that γ(u, v) 6= 0 and the coefficientsα, β, γ depend smoothly on u and v. Differentiate equation (4.31) with respect to v toobtain that α and γ are locally independent of v. Differentiate it with respect to uto obtain that β and γ are locally independent of u. Since the set of pairs of linearlyindependent vectors in V is connected, it follows that there exist functions α, β : V → Rand a constant c ∈ R \ 0 such that

u×′ v = α(u)v + β(v)u+ cu× v

for all pairs of linearly independent vectors u, v ∈ V . Interchange u and v to obtainβ(v) = −α(v) for all v ∈ V . Since the function V → V : u 7→ u×′ v is linear for all v ∈ Vit follows that α : V → R is linear. This proves Step 2.

Step 3. Let α and c be as in Step 2. Then α = 0 and 〈u, v〉′ = c2〈u, v〉 for all u, v ∈ V .

Fix a vector u ∈ V \ 0 and choose a vector v ∈ V such that u and v are linearly

independent. Then u × (u × v) = 〈u, v〉u − |u|2v by Lemma 2.9. Hence, it follows fromStep 2 that

〈u, v〉′u− |u|′2v = u×′ (u×′ v)

= u×′ (α(u)v − α(v)u+ cu× v)

= α(u)u×′ v + cu×′ (u× v)

= α(u)(α(u)v − α(v)u+ cu× v

)+ c

(α(u)u× v − α(u× v)u+ cu× (u× v)

)= α(u)

(α(u)v − α(v)u+ cu× v

)+ c

(α(u)u× v − α(u× v)u+ c〈u, v〉u− c|u|2v

)=(c2〈u, v〉 − cα(u× v)− α(u)α(v)

)u

+(α(u)2 − c2|u|2

)v + 2cα(u)u× v.

Since u, v, and u× v are linearly independent, it follows that

α(u) = 0, |u|′2 = c2|u|2 − α(u)2.

20


Since u ∈ V \ 0 was chosen arbitrarily, it follows that α(u) = 0 and

〈u, v〉′ = c2〈u, v〉, u×′ v = cu× vfor all u, v ∈ V . This proves Step 3.

Step 4. φ′ = ±φ.

It follows from Step 2 and Step 3 that

φ′(u, v, w) = 〈u×′ v, w〉′ = c3〈u× v, w〉 = c3φ(u, v, w)

for all u, v, w ∈ V , and so ψ = ψ′ = c4ψ by equation (4.9). Hence c = ±1 and this provesTheorem 4.30.

The next theorem follows a suggestion by Donaldson for characterizing coassociativecalibrations in terms of their dual 3–forms.

Theorem 4.32. Let V be a 7–dimensional vector space over the reals and let ψ ∈ Λ4V ∗.Then the following are equivalent:

(i) There exists a nondegenerate 3–form φ ∈ Λ3V ∗ and a number ε = ±1 such that εψis the coassociative calibration of (V, φ).

(ii) If α, β ∈ V ∗ are linearly independent, then there exists a 1–form γ ∈ V ∗ such thatα ∧ β ∧ γ ∧ ψ 6= 0.

Proof. That (i) implies (ii) follows from equation (4.38) in Lemma 4.37 below. To provethe converse, assume (ii) and fix any volume form σ ∈ Λ7V ∗. Define the 3–form Φ on thedual space V ∗ by

Φ(α, β, γ) :=α ∧ β ∧ γ ∧ ψ

σfor α, β, γ ∈ V ∗. (4.33)

This 3–form is nondegenerate by (ii). Denote the corresponding coassociative calibra-tion by Ψ : V ∗ × V ∗ × V ∗ × V ∗ → R and let 〈·, ·〉V ∗ be the inner product on V ∗ deter-mined by Φ. Let κ : V → V ∗ be the isomorphism induced by this inner product, soα(u) = 〈α, κ(u)〉V ∗ for α ∈ V ∗ and u ∈ V . Let 〈·, ·〉V be the pullback under κ of the innerproduct on V ∗. Then φ := κ∗Φ ∈ Λ3V ∗ is a nondegenerate 3–form compatible with theinner product and the volume form

vol := 17κ∗Φ ∧ κ∗Ψ.

By equation (4.33),φ(u, v, w)σ = κ(u) ∧ κ(v) ∧ κ(w) ∧ ψ (4.34)

for all u, v, w ∈ V . Choose λ > 0 and ε = ±1 such that

vol = ελ−4/3σ. (4.35)

Replace σ by σλ := λσ in (4.33) to obtain Φλ = λ−1Φ. Its coassociative calibration isΨλ = λ−4/3Ψ, the inner product on V ∗ induced by Φλ is 〈·, ·〉V ∗,λ = λ−2/3〈·, ·〉V ∗ , and

the isomorphism κλ : V → V ∗ is κλ = λ2/3κ. Hence ,

φλ := κ∗λΦλ = λφ, ψλ := κ∗λΨλ = λ4/3κ∗Ψ.

21


By (4.35) this implies

volλ := 17φλ ∧ ψλ = λ7/3vol = ελσ = εσλ.

Multiply both sides in equation (4.34) by ελ2 to obtain

φλ(u, v, w)εσλ = κλ(u) ∧ κλ(v) ∧ κλ(w) ∧ εψ

Since εσλ = volλ, it follows from (4.38) below that the same equation holds with εψreplaced by ψλ. Thus εψ = ψλ is the associative calibration of φλ. (Here ε is independentof the choice of σ.) This proves Theorem 4.32.

Remark 4.36. We can interpret Theorem 4.32 in the spirit of Remark 3.10. In the notationof Remark 3.10, if V is an oriented n–dimensional vector space with an inner product g,then the Hodge ∗–operator ∗ : ΛkV ∗ → Λn−kV ∗ can be defined as

∗α = (i−1g )∗α⊗ volg ∈ ΛkV ⊗ ΛnV ∗ = Λn−kV ∗.

If V is a 7–dimensional vector space and ψ ∈ Λ4V ∗, then we can equivalently thinkof it as a 3–form φ∗ on V ∗ with values in Λ7V ∗ since Λ4V ∗ = Λ3V ⊗ Λ7V ∗. Define asymmetric bilinear form G∗ : V ∗ × V ∗ → (Λ7V ∗)2 by

G∗(α, β) :=1

6i(α)φ∗ ∧ i(β)φ∗ ∧ φ∗ for α, β ∈ V ∗.

Condition (ii) in Theorem 4.32 is equivalent to iG∗ : V ∗ → V ⊗ (Λ7V ∗)2 being an isomor-phism. Note that det iG∗ ∈ (Λ7V ∗)12. After picking an orientation we define a positiveroot (det iG∗)

1/12 ∈ Λ7V ∗. Define a volume form on V and an inner product on V ∗ by

volg := (det(iG∗))1/12 and g∗ :=

G∗

vol2g.

A moment’s thought shows that volg is the volume form associated with the dual inner

product g and the chosen orientation on V . Further, the 3–form φ :=(ig)∗φ∗

volg∈ Λ3V ∗

satisfies 16 i(u)φ ∧ i(v)φ ∧ φ = g(u, v) volg. and ∗φ = ψ.

The next lemma summarizes some useful identities that will be needed throughout.The first of these has already been used in the proof of Theorem 4.32. Assume that Vis a 7–dimensional oriented real Hilbert space equipped with a compatible cross product,φ ∈ Λ3V ∗ is the associative calibration, and ψ := ∗φ ∈ Λ4V ∗ is the coassociative calibra-tion of (V, φ).

22


Lemma 4.37. The following hold for all u, v, w, x ∈ V and all ω ∈ Λ2V ∗:

ψ ∧ u∗ ∧ v∗ ∧ w∗ = φ(u, v, w)vol, (4.38)

φ ∧ u∗ ∧ v∗ ∧ w∗ ∧ x∗ = ψ(u, v, w, x)vol, (4.39)

ι(u)ψ ∧ v∗ ∧ ι(v)ψ = 0, (4.40)

∗(ψ ∧ u∗) = ι(u)φ, (4.41)

∗(φ ∧ u∗) = ι(u)ψ, (4.42)

|ι(u)φ|2 = 3|u|2, (4.43)

|ι(u)ψ|2 = 4|u|2, (4.44)

φ ∧ ι(u)φ = 2ψ ∧ u∗, (4.45)

φ ∧ ι(u)ψ = −4ι(u)vol, (4.46)

ψ ∧ ι(u)φ = 3ι(u)vol, (4.47)

ψ ∧ ι(u)ψ = 0, (4.48)

∗(φ ∧ ι(u)φ) = 2ι(u)φ, (4.49)

∗(φ ∧ ι(u)ψ) = −4u∗, (4.50)

∗(ψ ∧ ι(u)φ) = 3u∗, (4.51)

∗(ψ ∧ ∗(ψ ∧ ι(u)φ)) = 3ι(u)φ, (4.52)

ι(u)φ ∧ ∗ι(v)φ = 3〈u, v〉vol, (4.53)

u∗ ∧ v∗ = ι(u× v)φ− ι(v)ι(u)ψ, (4.54)

u∗ ∧ v∗ ∧ ∗ι(u× v)φ = |u× v|2vol, (4.55)

u∗ ∧ v∗ ∧ ι(u)φ ∧ ι(v)ψ = 2|u× v|2vol, (4.56)

ψ ∧ u∗ ∧ v∗ = ι(u× v)vol, (4.57)

φ ∧ u∗ ∧ v∗ ∧ w∗ = ι([u, v, w])vol, (4.58)

φ ∧ u∗ ∧ v∗ = ∗ι(v)ι(u)ψ, (4.59)

∗(ψ ∧ ∗(ψ ∧ ω)) = ω + ∗(φ ∧ ω), (4.60)

∗(φ ∧ ∗(φ ∧ ω)) = 2ω + ∗(φ ∧ ω). (4.61)

Proof. It is a general fact about alternating k–forms on a finite-dimensional Hilbert spaceV that 〈u∗1∧· · ·∧u∗k, α〉 = α(u1, . . . , uk) for all ui ∈ V and all α ∈ ΛkV ∗. This proves (4.38)and (4.39). Equations (4.41) and (4.42) follow from (4.15) in Remark 4.14.

To prove equations (4.40) and (4.43)–(4.47) assume without loss of generality that u, vare orthonormal. By Theorem 3.2 assume that V = R7 with u = e1 and v = e2, and thatφ and ψ are as in (4.12) and (4.13), i.e.,

φ = φ0 = e123 − e145 − e167 − e246 + e257 − e347 − e356,

ψ = ψ0 = −e1247 − e1256 + e1346 − e1357 − e2345 − e2367 + e4567.(4.62)

23


Then

ι(u)φ = e23 − e45 − e67,

ι(u)ψ = −e247 − e256 + e346 − e357,

v∗ ∧ ι(v)ψ = −e1247 − e1256 + e2345 − e2367.

(4.63)

Equation (4.40) follows by multipying the last two sums, and (4.43) and (4.44) follow byexamining the first two sums. Moreover, by (4.62) and (4.63),

φ ∧ ι(u)φ = −2e12345 − 2e12367 + 2e14567 = 2 ∗ι(u)φ = 2u∗ ∧ ψ.This proves (4.45). By (4.62) and (4.63) we also have ψ ∧ ι(u)φ = 3e234567 andφ ∧ ι(u)ψ = −4e234567. This proves (4.46) and (4.47).

Equation (4.48) follows by contracting u with the 8–form ψ∧ψ = 0. Equations (4.49)–(4.51) follow from (4.45)–(4.47) and the fact that ∗u∗ = ι(u)vol and ∗(u∗ ∧ ψ) = ι(u)φby (4.41). To prove equation (4.52) take the exterior product of equation (4.51) with ψand then use (4.41) to obtain

ψ ∧ ∗(ψ ∧ ι(u)φ) = ψ ∧ 3u∗ = 3 ∗ι(u)φ.

Equation (4.53) follows from (4.43) and the fact that the left hand side in (4.53) issymmetric in u and v. Equation (4.54) is equivalent to (4.20) in the proof of Lemma 4.8.To prove equation (4.55) choose w := u× v in (4.38) to obtain

|u× v|2vol = u∗ ∧ v∗ ∧ (u× v)∗ ∧ ψ = u∗ ∧ v∗ ∧ ∗ι(u× v)φ.

Here the last equation follows from (4.41). To prove (4.56) we compute

u∗ ∧ v∗ ∧ ι(u)φ ∧ ι(v)ψ

= −ι(v)(u∗ ∧ v∗ ∧ ι(u)φ

)∧ ψ

= −〈u, v〉v∗ ∧ ι(u)φ ∧ ψ + |v|2u∗ ∧ ι(u)φ ∧ ψ − u∗ ∧ v∗ ∧ (u× v)∗ ∧ ψ

= −〈u, v〉ι(u)φ ∧ ∗ι(v)φ+ |v|2ι(u)φ ∧ ∗ι(u)φ− u∗ ∧ v∗ ∧ ∗ι(u× v)φ

= 2|u× v|2vol.

Here the second step uses the identity ι(v)ι(u)φ = φ(u, v, ·) = (u × v)∗, the third stepfollows from (4.41), and the last step follows from (4.45) and (4.55).

To prove equation (4.57) take the exterior product with a 1–form w∗ and use equa-tion (4.38) to obtain(

ψ ∧ u∗ ∧ v∗)∧ w∗ = φ(u, v, w)vol = 〈u× v, w〉vol

= (∗(u× v)∗) ∧ w∗ = (ι(u× v)vol) ∧ w∗.

To prove equation (4.58) take the exterior product with a 1–form x∗ and use equa-tion (4.39) to obtain(

φ ∧ u∗ ∧ v∗ ∧ w∗)∧ x∗ = ψ(u, v, w, x)vol = 〈[u, v, w], x〉vol

= (∗[u, v, w]∗) ∧ x∗ = (ι([u, v, w])vol) ∧ x∗.

24


To prove equation (4.59) take the exterior product with w∗ ∧ x∗ for w, x ∈ V and useequation (4.57) to obtain(

φ ∧ u∗ ∧ v∗)∧ (w∗ ∧ x∗) = ψ(u, v, w, x)vol

= 〈ι(v)ι(u)ψ,w∗ ∧ x∗〉vol

= (∗ι(v)ι(u)ψ) ∧ (w∗ ∧ x∗).

Since Λ2V ∗ has a basis of 2–forms of the form w∗ ∧ x∗, this proves (4.59).To prove equations (4.60) and (4.61) it suffices to assume

ω = u∗ ∧ v∗

for u, v ∈ V . Then it follows from (4.54) and (4.59) that

ι(u× v)φ = u∗ ∧ v∗ + ι(v)ι(u)ψ

= u∗ ∧ v∗ + ∗(u∗ ∧ v∗ ∧ φ)

= ω + ∗(φ ∧ ω).

(4.64)

Moroever, ∗(ψ ∧ ω) = (u× v)∗ by (4.57). Hence, by (4.41) and (4.64),

∗(ψ ∧ ∗

(ψ ∧ ω

))= ∗(ψ ∧ (u× v)∗

)= ι(u× v)φ

= ω + ∗(φ ∧ ω).

This proves equation (4.60). Moreover, by (4.49) and (4.64),

∗(φ ∧ ∗

(φ ∧ ω

))= ∗(φ ∧

(ι(u× v)φ− ω

))= ∗(φ ∧ ι(u× v)φ

)− ∗(φ ∧ ω

)= 2ι(u× v)φ− ∗

(φ ∧ ω

)= 2ω + ∗(φ ∧ ω).

This proves equation (4.61) and Lemma 4.37.

5. Normed algebras

Definition 5.1. A normed algebra consists of a finite dimensional real Hilbert spaceW , a bilinear map

W ×W →W : (u, v) 7→ uv,

(called the product), and a unit vector 1 ∈W (called the unit), satisfying

1u = u1 = u

and

|uv| = |u||v| (5.2)

for all u, v ∈W .

25


When W is a normed algebra it is convenient to identify the real numbers with asubspace of W via multiplication with the unit 1. Thus, for u ∈ W and λ ∈ R, we writeu+ λ instead of u+ λ1. Define an involution W → W : u 7→ u (called conjugation) by1 := 1 and u := −u for u ∈ 1⊥. Thus

u := 2〈u, 1〉 − u. (5.3)

We think of R ⊂ W as the real part of W and of its orthogonal complement as theimaginary part. The real and imaginary parts of u ∈W will be denoted by Reu := 〈u, 1〉and Imu := u− 〈u, 1〉.

Theorem 5.4. Normed algebras and vector spaces with cross products are related asfollows.

(i) If W is a normed algebra, then V := 1⊥ is equipped with a cross productV × V → V : (u, v) 7→ u× v defined by

u× v := uv + 〈u, v〉 (5.5)

for u, v ∈ 1⊥.(ii) If V is a finite dimensional Hilbert space equipped with a cross product, then

W := R⊕ V is a normed algebra with

uv := u0v0 − 〈u1, v1〉+ u0v1 + v0u1 + u1 × v1 (5.6)

for u = u0 + u1, v = v0 + v1 ∈ R ⊕ V . Here we identify a real number λ with thepair (λ, 0) ∈ R⊕ V and a vector v ∈ V with the pair (0, v) ∈ R⊕ V .

These constructions are inverses of each other. In particular, a normed algebra hasdimension 1, 2, 4, or 8 and is isomorphic to R, C, H, or O.

Proof. See page 27.

Lemma 5.7. Let W be a normed algebra. Then the following hold:

(i) For all u, v, w ∈W we have

〈uv,w〉 = 〈v, uw〉, 〈uv,w〉 = 〈u,wv〉. (5.8)

(ii) For all u, v ∈W we have

uu = |u|2, uv + vu = 2〈u, v〉. (5.9)

(iii) For all u, v ∈W we have

〈u, v〉 = 〈u, v〉, uv = vu. (5.10)

(iv) For all u, v, w ∈W we have

u(vw) + v(uw) = 2〈u, v〉w, (uv)w + (uw)v = 2〈v, w〉u. (5.11)

26


Proof. We prove (i). The first equation in (5.8) is obvious when u is a real multipleof 1. Hence, it suffices to assume that u is orthogonal to 1. Expanding the identities|uv + uw|2 = |u|2|v + w|2 and |uv + wv|2 = |u+ w|2|v|2 we obtain the equations

〈uv, uw〉 = |u|2〈v, w〉, 〈uv,wv〉 = 〈u,w〉|v|2. (5.12)

If u is orthogonal to 1, the first equation in (5.12) gives

〈uv,w〉+ 〈v, uw〉 = 〈(1 + u)v, (1 + u)w〉 − (1 + |u|2)〈v, w〉 = 0.

Since u = −u for u ∈ 1⊥, this proves the first equation in (5.8). The proof of the secondequation is similar.

We prove (ii). Using the second equation in (5.8) with v = u we obtain

〈uu, w〉 = 〈u,wu〉 = 〈1, w〉|u|2.

Here we have used the second equation in (5.12). This implies uu = |u|2 for every u ∈W .Replacing u by u+ v gives uv + vu = 2〈u, v〉. This proves (5.9).

We prove (iii). That conjugation is an isometry follows immediately from the definition.Using (5.9) with v replaced by v we obtain

vu = 2〈u, v〉 − uv = 2〈uv, 1〉 − uv = uv.

Here the second equation follows from (5.8). This proves (5.10).We prove (iv). For all u,w ∈W we have

〈u(uw), w〉 = |uw|2 = |u|2|w|2 = |u|2|w|2. (5.13)

Since the operator w 7→ u(uw) is self-adjoint, by (5.8), this shows that u(uw) = |u|2w forall u,w ∈ W . Replacing u by u + v we obtain the first equation in (5.11). The proof ofthe second equation is similar.

Proof of Theorem 5.4. Let W be a normed algebra. It follows from (5.8) that

〈u, v〉 = −〈uv, 1〉

and, hence, u × v := uv + 〈u, v〉 ∈ 1⊥ for all u, v ∈ 1⊥. We write an element of W asu = u0 + u1 with u0 := 〈u, 1〉 ∈ R and u1 := u − 〈u, 1〉 ∈ V = 1⊥. For u, v ∈ W wecompute

|u|2|v|2 − |uv|2 =(u2

0 + |u1|2)(

v20 + |v1|2

)− (u0v0 − 〈u1, v1〉)2

− |u0v1 + v0u1 + u1 × v1|2

= u20|v1|2 + v2

0 |u1|2 + 2u0v0〈u1, v1〉+ |u1|2|v1|2 − 〈u1, v1〉2

− |u0v1 + v0u1|2 − |u1 × v1|2 − 2〈u0v1 + v0u1, u1 × v1〉

=|u1|2|v1|2 − 〈u1, v1〉2 − |u1 × v1|2

− 2u0〈v1, u1 × v1〉 − 2v0〈u1, u1 × v1〉.

27


The right hand side vanishes for all u and v if and only if the product on V satisfies (2.3)and (2.4). Hence, (5.5) defines a cross product on V and the product can obviously berecovered from the cross product via (5.6). Conversely, the same argument shows that, ifV is equipped with a cross product, the formula (5.6) defines a normed algebra structureon W := R⊕ V . Moreover, by Theorem 2.5, V has dimension 0, 1, 3, or 7. This provesTheorem 5.4.

Remark 5.14. If W is a normed algebra and the cross product on V := 1⊥ is definedby (5.5), then the commutator of two elements u, v ∈W is given by

[u, v] := uv − vu = 2u1 × v1. (5.15)

In particular, the product on W is commutative in dimensions 1 and 2 and is not com-mutative in dimensions 4 and 8.

Remark 5.16. Let W be a normed algebra of dimension 4 or 8. Then V := 1⊥ has anatural orientation determined by Lemma 2.12 or Lemma 3.4, respectively, in dimensions3 and 7. We orient W as R⊕ V .

Remark 5.17. If W is a normed algebra and the cross product on V := 1⊥ is definedby (5.5), then the associator bracket on V is related to the product on W by

(uv)w − u(vw) = 2[u1, v1, w1] (5.18)

for all u, v, w ∈ W . Thus W is an associative algebra in dimensions 1, 2, 4 and is notassociative in dimension 8. The formula (5.18) is the reason for the term associatorbracket. Many authors actually define the associator bracket as the left hand side ofequation (5.18) (see for example [HL82]).

To prove (5.18), we observe that the associator bracket on V can be written in theform

2[u, v, w] = 2(u× v)× w + 2〈v, w〉u− 2〈u,w〉v= (u× v)× w − u× (v × w) + 〈v, w〉u− 〈u, v〉w

(5.19)

for u, v, w ∈ V . Here the first equation follows from (4.1) and the second equation followsfrom (2.11). For u, v, w ∈ V we compute

(uv)w − u(vw) = (−〈u, v〉+ u× v)w − u(−〈v, w〉+ v × w)

= (u× v)× w − u× (v × w)− 〈u, v〉w + 〈v, w〉u= 2[u, v, w].

Here the first equation follows from the definition of the cross product in (5.5), the secondequation follows by applying (5.5) again and using (2.7), and the last equation followsfrom (5.19). Now, if any of the factors u, v, w is a real number, the term on the leftvanishes. Hence, real parts can be added to the vectors without changing the expression.

28


Theorem 5.20. Let W be an 8–dimensional normed algebra.

(i) The map W 3 →W : (u, v, w) 7→ u× v × w defined by

u× v × w := 12

((uv)w − (wv)u

)(5.21)

(called the triple cross product of W ) is alternating and satisfies

〈x, u× v × w〉+ 〈u× v × x,w〉 = 0, (5.22)

|u× v × w| = |u ∧ v ∧ w|, (5.23)

for all u, v, w, x ∈W and

〈e× u× v, e× w × x〉 = −|e|2〈u× v × w, x〉 (5.24)

whenever e, u, v, w, x ∈W are orthonormal.(ii) The map Φ : W 4 → R defined by

Φ(x, u, v, w) := 〈x, u× v × w〉

(called the Cayley calibration of W ) is an alternating 4–form. Moreover, Φ isself-dual, i.e.,

Φ = ∗Φ, (5.25)

where ∗ : ΛkW ∗ → Λ8−kW ∗ denotes the Hodge ∗–operator associated to the innerproduct and the orientation of Remark 5.16.

(iii) Let V := 1⊥ with the cross product defined by (5.5) and the associator bracket [·, ·, ·]defined by (4.1). Let φ ∈ Λ3V ∗ and ψ ∈ Λ4V ∗ be the associative and coassocia-tive calibrations of V defined by (2.8) and (4.9), respectively. Then the triple crossproduct (5.21) of u, v, w ∈W can be expressed as

u× v × w = φ(u1, v1, w1)− [u1, v1, w1]

− u0(v1 × w1)− v0(w1 × u1)− w0(u1 × v1)(5.26)

and the Cayley calibration is given by

Φ = 1∗ ∧ φ+ ψ. (5.27)

(iv) For all u, v ∈W we have

uv = u× 1× v + 〈u, 1〉v + 〈v, 1〉u− 〈u, v〉. (5.28)

Remark 5.29. There is a choice involved in the definition of the triple cross productin (5.21). An alternative formula is

(u, v, w) 7→ 12

(u(vw)− w(vu)

).

This map also satisfies (5.22) and (5.23). However, it satisfies (5.24) with the minus signchanged to plus and the resulting Cayley calibration is given by Φ = 1∗ ∧ φ − ψ and isanti-self-dual. Equation (5.28) remains unchanged.

29


Proof of Theorem 5.20. Let W ×W ×W → W : (u, v, w) 7→ u × v × w be the trilinearmap defined by (5.21). We prove that this map satisfies (5.26). To see this, fix threevectors u, v, w ∈W . Then, by (5.15), we have

vw − wv = −2v1 × w1, uw − wu = −2w1 × u1, uv − vu = −2u1 × v1.

Multiplying these expressions by u0, v0, w0, respectively, we obtain (twice) the last threeexpressions on the right in (5.26). Thus it suffices to assume u, v, w ∈ V . Then we obtain

2u× v × w = (uv)w − (wv)u

= −(uv)w + (wv)u

= −(−〈u, v〉+ u× v)w + (−〈w, v〉+ w × v)u

= 〈u× v, w〉+ 〈u, v〉w − (u× v)× w− 〈w × v, u〉 − 〈w, v〉u+ (w × v)× u

= 2φ(u, v, w)− 2[u, v, w].

Here the third and fourth equations follow from (5.5), and the last equation followsfrom (2.8) and (5.19). This proves that the formulas (5.21) and (5.26) agree.

We prove (i). By (5.26) we have

〈x, u× v × w〉 = x0φ(u1, v1, w1) + ψ(x1, u1, v1, w1)

− u0φ(x1, v1, w1)− v0φ(x1, w1, u1)

− w0φ(x1, u1, v1)

(5.30)

for x, u, v, w ∈W . Here we have used φ(u1, v1, w1) = 〈u1, v1 × w1〉 and

−〈x1, [u1, v1, w1]〉 = −ψ(u1, v1, w1, x1) = ψ(x1, u1, v1, w1).

It follows from the alternating properties of φ and ψ that the right hand side of (5.30)is an alternating 4–form. Hence, the map (5.21) is alternating and satisfies (5.22). Foru, v, w ∈ V = 1⊥ equation (5.23) follows from Lemma 4.4. In general, if u, v, w ∈ W arepairwise orthogonal, it follows from (5.9) and (5.11) that

(uv)w = −(uw)v = (wu)v = −(wv)u.

This shows that

〈u, v〉 = 〈v, w〉 = 〈w, u〉 = 0 =⇒ u× v × w = u(vw) (5.31)

and, hence, by (5.2), we have |u × v × w| = |u ∧ v ∧ w| in the orthogonal case. Thisequation continues to hold in general by Gram–Schmidt. This proves that the triple crossproduct satisfies (5.23).

30


We prove (5.24). The second equation in (5.11) asserts that (yz)z = |z|2y for ally, z ∈W . Hence, by (5.31), we have

〈e× u× v, e× w × x〉 = 〈u× v × e, w × x× e〉= 〈(uv)e, (wx)e〉= 〈uv, ((wx)e)e〉

= |e|2〈uv, wx〉

= |e|2〈(uv)x,w〉

= |e|2〈u× v × x,w〉

= −|e|2〈x, u× v × w〉

whenever e, u, v, w, x ∈ W are pairwise orthogonal. Thus the triple cross product (5.21)satisfies (5.24). This proves (i).

We prove (ii) and (iii). That Φ is a 4–form follows from (i). That it satisfies equa-tion (5.27) follows directly from the definition of Φ and equation (5.30). That Φ is self-dual with respect to the orientation of Remark 5.16 follows from (5.27) and Lemma 4.8.Equation (5.26) was proved above.

We prove (iv). By (5.15) and (5.21), we have

u1 × v1 = 12

(uv − vu

)= u× 1× v.

Hence, it follows from (5.6) that

uv = u0v0 − 〈u1, v1〉+ u0v1 + v0u1 + u1 × v1

= −u0v0 − 〈u1, v1〉+ u0v + v0u+ u1 × v1

= −〈u, v〉+ 〈u, 1〉v + 〈v, 1〉u+ u× 1× v.

This proves (5.28) and Theorem 5.20.

Example 5.32. If W = R8 = R ⊕ R7 with coordinates x0, x1, . . . , x7 and the crossproduct of Example 2.15 on R7, then the associated Cayley calibration is given by

Φ0 = e0123 − e0145 − e0167 − e0246 + e0257 − e0347 − e0356

+ e4567 − e2367 − e2345 − e1357 + e1346 − e1256 − e1247.

Thus

Φ0 ∧ Φ0 = 14vol.

(See the proof of Lemma 4.8.)

Definition 5.33. Let W be an 8–dimensional normed algebra. The fourfold crossproduct on W is the alternating multi-linear map W 4 →W : (u, v, w, x) 7→ u×v×w×xdefined by

4x× u× v × w := (u× v × w)x− (v × w × x)u+ (w × x× u)v − (x× u× v)w. (5.34)

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Theorem 5.35. Let W be an 8–dimensional normed algebra with triple cross prod-uct (5.21), Cayley calibration Φ ∈ Λ4W ∗, and fourfold cross product (5.34). Then, for allx, u, v, w ∈W , we have

|x× u× v × w| = |x ∧ u ∧ v ∧ w| (5.36)

and

Re (x× u× v × w) = Φ(x, u, v, w),

Im (x× u× v × w) = [x1, u1, v1, w1]− x0[u1, v1, w1]

+ u0[v1, w1, x1]− v0[w1, x1, u1]

+ w0[x1, u1, v1],

(5.37)

where the last five terms use the associator and coassociator brackets on V := 1⊥ definedby (4.1) and (4.22). In particular,

Φ(x, u, v, w)2 + |Im (x× u× v × w)|2 = |x ∧ u ∧ v ∧ w|2. (5.38)

Proof. That the fourfold cross product is alternating is obvious from the definition andthe alternating property of the triple cross product. We prove that it satisfies (5.36). Forthis it suffices to assume that u, v, w, x are pairwise orthogonal. Then u× v×w = (uv)wand hence

(u× v × w)x = ((uv)w)x = −((uv)x)w = −(u× v × x)w.

Here we have used (5.9) and (5.11). Using the alternating property of the triple crossproduct we obtain that the four summands in (5.34) agree in the orthogonal case. Hence,

x× u× v × w = ((uv)w)x

and so equation (5.36) follows from (5.2).We prove (5.37). Since u× 1× v = u1 × v1, we have

1× u× v × w = 14

(u× v × w + (v1 × w1)u+ (w1 × u1)v + (u1 × v1)w

)= 1

4

(u× v × w + u0(v1 × w1) + v0(w1 × u1) + w0(u1 × v1)

)+ 1

4

(〈v1 × w1, u1〉+ 〈w1 × u1, v1〉+ 〈u1 × v1, w1〉

)− 1

4

((v1 × w1)× u1 + (w1 × u1)× v1 + (u1 × v1)× w1

)= φ(u1, v1, w1)− [u1, v1, w1].

The last equation follows from (5.26) and the definition of the associator bracket in (4.1).This proves (5.37) in the case x1 = 0. Using the alternating property we may now assumethat x, u, v, w ∈ V := 1⊥. If x, u, v, w are orthogonal to 1 it follows from (5.26) that

u× v × w = φ(u, v, w)− [u, v, w]

and

Φ(x, u, v, w) = −〈x, [u, v, w]〉 = ψ(x, u, v, w).

32


Moreover, x = −x and similarly for u, v, w. Hence,

4x× u× v × w= −(u× v × w)x+ (v × w × x)u− (w × x× u)v + (x× u× v)w

= [u, v, w]x− [v, w, x]u+ [w, x, u]v − [x, u, v]w

− φ(u, v, w)x+ φ(v, w, x)u− φ(w, x, u)v + φ(x, u, v)w

= −〈[u, v, w], x〉+ 〈[v, w, x], u〉 − 〈[w, x, u], v〉+ 〈[x, u, v], w〉+ [u, v, w]× x− [v, w, x]× u+ [w, x, u]× v − [x, u, v]× w− φ(u, v, w)x+ φ(v, w, x)u− φ(w, x, u)v + φ(x, u, v)w

= −4ψ(u, v, w, x)− 4[u, v, w, x]

= 4Φ(x, u, v, w) + 4[x, u, v, w].

Here the last but one equation follows from Lemma 4.21. Thus we have proved (5.37)and Theorem 5.35.

6. Triple cross products

In this section we show how to recover the normed algebra structure on W from thetriple cross product. In fact we shall see that every unit vector in W can be used as aunit for the algebra structure. We assume throughout that W is a finite dimensional realHilbert space.

Definition 6.1. An alternating multi-linear map

W ×W ×W →W : (u, v, w) 7→ u× v × w (6.2)

is called a triple cross product if it satisfies

〈u× v × w, u〉 = 〈u× v × w, v〉 = 〈u× v × w,w〉 = 0, (6.3)

|u× v × w| = |u ∧ v ∧ w| (6.4)

for all u, v, w ∈W .

A multi-linear map (6.2) that satisfies (6.4) also satisfies u × v × w = 0 wheneveru, v, w ∈W are linearly dependent, and hence is necessarily alternating.

Lemma 6.5. Let (6.2) be an alternating multi-linear map. Then (6.3) holds if and onlyif, for all x, u, v, w ∈W , we have

〈x, u× v × w〉+ 〈u× v × x,w〉 = 0. (6.6)

Proof. If (6.6) holds, then (6.3) follows directly from the alternating property of themap (6.2). To prove the converse, expand the expression 〈u × v × (w + x), w + x〉 anduse (6.3) to obtain (6.6).

33


Lemma 6.7. Let (6.2) be an alternating multi-linear map satisfying (6.3). Then equa-tion (6.4) holds if and only if, for all u, v, w ∈W , we have

u× v × (u× v × w) + |u ∧ v|2w

=(|v|2〈u,w〉 − 〈u, v〉〈v, w〉

)u+

(|u|2〈v, w〉 − 〈v, u〉〈u,w〉

)v.

(6.8)

Proof. If (6.8) holds and w is orthogonal to u and v, then

u× v × (u× v × w) = −|u ∧ v|2w.

Taking the inner product with w and using (6.6) we obtain (6.4) under the assumption〈u,w〉 = 〈v, w〉 = 0. Since both sides of equation (6.4) remain unchanged if we add to wa linear combination of u and v, this proves that (6.8) implies (6.4).

To prove the converse we assume (6.4). If w is orthogonal to u and v, we have

|u× v × w|2 = |u ∧ v|2|w|2.

Replacing w by w + x we obtain

w, x ∈ u⊥ ∩ v⊥ =⇒ 〈u× v × w, u× v × x〉 = |u ∧ v|2〈w, x〉. (6.9)

Using (6.6) we obtain (6.8) for every vector w ∈ u⊥ ∩ v⊥. Replacing a general vector wby its projection onto the orthogonal complement of the subspace spanned by u and v wededuce that (6.8) holds in general. This proves Lemma 6.7.

Let (6.2) be a triple cross product. If e ∈ W is a unit vector, then the subspaceVe := e⊥ carries a cross product (u, v) 7→ u×e v defined by u×e v := u× e× v. Hence, byTheorem 2.5, the dimension of Ve is 0, 1, 3, or 7.

It follows that the dimension of W is 0, 1, 2, 4, or 8.

Lemma 6.10. Assume dimW = 8 and let (6.2) be a triple cross product. Then there isa number ε ∈ ±1 such that

e× u× (e× v × w) = ε|e|2u× v × w (6.11)

whenever e, u, v ∈ W are pairwise orthogonal and w ∈ W is orthogonal to e, u, v, ande× u× v.

Proof. It suffices to assume that the vectors e, u, v ∈ W are orthonormal. Then thesubspace

H := span(e, u, v, e× u× v)⊥

has dimension four. It follows from (6.6) and (6.9) that the formulas

Iw := e× u× w, Jw := e× v × w, Kw := u× v × w,

define endomorphisms I, J,K of H. Moreover, by (6.6), these operators are skew adjointand, by (6.9), they are complex structures on H. It follows also from (6.9) that

e× x× (e× x× w) = −|x|2w

34


whenever e, x, w are pairwise orthogonal and |e| = 1. Assuming w ∈ H and using thisidentity with x = u+ v we obtain IJ + JI = 0. This implies that the automorphisms ofH of the form aI + bJ + cIJ with a2 + b2 + c2 = 1 belong to the space J of orthogonalcomplex structures on H. They form one of the two components of J and K belongs tothis component because it anticommutes with I and J . Hence, K = εIJ with ε = ±1.Since the space of orthonormal triples in W is connected, and the constant ε dependscontinuously on the triple e, u, v, we have proved (6.11) under the assumption that e, u, vare orthonormal and w is orthogonal to the vectors e, u, v, e× u× v. Hence, the assertionfollows by scaling. This proves Lemma 6.10.

Definition 6.12. Assume dimW = 8. A triple cross product (6.2) is called positive ifit satisfies (6.11) with ε = 1 and is called negative if it satisfies (6.11) with ε = −1.

Definition 6.13. Assume dimW = 8 and let (6.2) be a triple cross product. Then, byLemma 6.5, the map Φ : W ×W ×W ×W → R defined by

Φ(x, u, v, w) := 〈x, u× v × w〉 (6.14)

is an alternating 4–form. It is called the Cayley calibration of W .

Theorem 6.15. Assume dimW = 8 and let (6.2) be a triple cross product with Cayleycalibration Φ ∈ Λ4W ∗ given by (6.14). Let e ∈W be a unit vector.

(i) Define the map ψe : W 4 → R by

ψe(u, v, w, x) := 〈e× u× v, e× w × x〉−(〈u,w〉 − 〈u, e〉〈e, w〉

)(〈v, x〉 − 〈v, e〉〈e, x〉

)+(〈u, x〉 − 〈u, e〉〈e, x〉

)(〈v, w〉 − 〈v, e〉〈e, w〉

).

(6.16)

Then ψe ∈ Λ4W ∗ and

Φ = e∗ ∧ φe + εψe, φe := ι(e)Φ ∈ Λ3W ∗, (6.17)

where ε ∈ ±1 is as in Lemma 6.10.(ii) The subspace Ve := e⊥ carries a cross product

Ve × Ve → Ve : (u, v) 7→ u×e v := u× e× v, (6.18)

the restriction of φe to Ve is the associative calibration of (6.18), and the restrictionof ψe to Ve is the coassociative calibration of (6.18).

(iii) The space W is a normed algebra with unit e and multiplication and conjugationgiven by

uv := u× e× v + 〈u, e〉v + 〈v, e〉u− 〈u, v〉e, u := 2〈u, e〉e− u. (6.19)

If the triple cross product is positive, then (uv)w − (wv)u = 2u× v × w.

Proof. We prove (i). If the vectors e, u, v, w, x are pairwise orthogonal, then

〈e× u× x, e× v × w〉 = −ε|e|2〈x, u× v × w〉. (6.20)

35


To see this, take the inner product of (6.11) with x. Then it follows from (6.6) that (6.20)holds under the additional assumption that w is perpendicular to e × u × v. Since x isorthogonal to e, this additional condition can be dropped, as both sides of the equationremain unchanged if we add to w a multiple of e× u× v. Thus we have proved (6.20).

Now fix a unit vector e ∈ W . By definition, ψe is alternating in the first two and lasttwo arguments, and satisfies ψe(u, v, w, x) = ψe(w, x, u, v) for all u, v, w, x ∈W . By (6.4)we also have ψe(u, v, u, v) = 0. Expanding the identity ψe(u, v+x, u, v+x) = 0 we obtainψe(u, v, u, x) = 0 for all u, v, x ∈W . Using this identity with u replaced by u+ w gives

ψe(u, v, w, x) + ψe(w, v, u, x) = 0.

Hence, ψe is also skew-symmetric in the first and third argument and so is an al-ternating 4–form. To see that it satisfies (6.17) it suffices to show that εΦ and ψeagree on e⊥. Since they are both 4–forms, it suffices to show that they agree on ev-ery quadrupel of pairwise orthogonal vectors u, v, w, x ∈ e⊥. But in this case we haveψe(u, x, v, w) = −εΦ(x, u, v, w) = εΦ(u, x, v, w), by equation (6.20). This proves (i).

We prove (ii). That (6.18) is a cross product on Ve = e⊥ follows immediately from thedefinitions.

By (6.14) we have

〈u×e v, w〉 = Φ(w, u, e, v) = Φ(e, u, v, w) = φe(u, v, w)

for u, v, w ∈ Ve, and hence the restriction of φe to Ve is the associative calibration.Moreover, the associator bracket (4.1) on Ve is given by

[u, v, w]e = (u× e× v)× e× w + 〈v, w〉u− 〈u,w〉v.Hence, for all u, v, w, x ∈ Ve, we have

〈[u, v, w]e, x〉 = 〈e× w × (u× e× v), x〉+ 〈v, w〉〈u, x〉 − 〈u,w〉〈v, x〉= 〈e× u× v, e× w × x〉 − 〈u,w〉〈v, x〉+ 〈u, x〉〈v, w〉= ψe(u, v, w, x),

where the last equation follows from (6.16). Hence, the restriction of ψe to Ve is thecoassociative calibration and this proves (ii).

We prove (iii). That e is a unit follows directly from the definitions. To prove that thenorm of the product is equal to the product of the norms we observe that u × e × v isorthogonal to e, u, and v, by equation (6.6). Hence,

|uv|2 = |u× e× v + 〈u, e〉v + 〈v, e〉u− 〈u, v〉e|2

= |u× e× v|2 − 2〈v, e〉〈u, v〉〈v, e〉

+ 〈u, e〉2|v|2 + 〈v, e〉2|u|2 + 〈u, v〉2

= |u|2|v|2.

Here the last equality uses the fact that |u× e× v|2 = |u ∧ e ∧ v|2. Thus we have provedthat W is a normed algebra with unit e.

36


If the triple cross product (6.2) is positive, then ε = 1 and hence equation (6.17)asserts that Φ = e∗ ∧ φe + ψe. Hence, it follows from (5.27) in Theorem 5.20 that theCayley calibration Φe associated to the above normed algebra structure is equal to Φ.This implies that the given triple cross product (6.2) agrees with the triple cross productdefined by (5.21). This proves (iii) and Theorem 6.15.

Remark 6.21. Assume dimW = 8 and let (6.2) be a triple cross product with Cayleycalibration Φ ∈ Λ4W ∗ given by (6.14). Then, for every unit vector e ∈ W , the subspaceVe = e⊥ is oriented by Lemma 3.4 and Theorem 6.15. We orient W as the direct sumW = Re ⊕ Ve. This orientation is independent of the choice of the unit vector e. Withthis orientation we have e∗ ∧ φe = ∗ψe, by Theorem 6.15 (ii) and Lemma 4.8. Hence, itfollows from equation (6.17) in Theorem 6.15 (i) that Φ ∧Φ 6= 0. In fact, the triple crossproduct is positive if and only if Φ ∧ Φ > 0 with respect to our orientation and negativeif and only if Φ ∧ Φ < 0. In the positive case Φ is self-dual and in the negative case Φ isanti-self-dual.

Corollary 6.22. Assume dimW = 8 and let (6.2) be a triple cross product and let ε beas in Lemma 6.10. Then, for all e, u, v, w ∈W , we have

e× u× (e× v × w) = ε|e|2u× v × w − ε〈e, u× v × w〉e− ε〈e, u〉e× v × w− ε〈e, v〉e× w × u− ε〈e, w〉e× u× v

−(|e|2〈u, v〉 − 〈e, u〉〈e, v〉

)w

+(|e|2〈u,w〉 − 〈e, u〉〈e, w〉

)v

+(〈u, v〉〈e, w〉 − 〈u,w〉〈e, v〉

)e.

(6.23)

Proof. Both sides of the equation remain unchanged if we add to u, v, or w a multiple ofe. Hence, it suffices to prove (6.23) under the assumption that u, v, w are all orthogonalto e. Moreover, both sides of the equation are always orthogonal to e. Hence, it sufficesto prove that the inner products of both sides of (6.23) with every vector x ∈ e⊥ agree.It also suffices to assume |e| = 1. Thus we must prove that, if e ∈W is a unit vector andu, v, w, x ∈W are orthogonal to e, then we have

〈e× u× (e× v × w), x〉 = ε〈u× v × w, x〉 − 〈u, v〉〈w, x〉+ 〈u,w〉〈v, x〉

or equivalently

− 〈e× u× x, e× v × w〉+ 〈u, v〉〈x,w〉 − 〈u,w〉〈x, v〉 = ε〈x, u× v × w〉. (6.24)

The right hand side of (6.24) is εΦ(x, u, v, w) and, by (6.16), the left hand side of (6.24) is−ψe(u, x, v, w). Hence, equation (6.24) is equivalent to the assertion that the restrictionof ψe to e⊥ agrees with Φ. But this follows from equation (6.17) in Theorem 6.15. Thisproves Corollary 6.22.

37


Lemma 6.25. Assume dimW = 8 and let (6.2) be a triple cross product with Cayleycalibration Φ ∈ Λ4W ∗ given by (6.14). Let H ⊂ W be a 4–dimensional linear subspace.Then the following are equivalent:

(i) If u, v, w ∈ H, then u× v × w ∈ H.(ii) If u, v ∈ H and w ∈ H⊥, then u× v × w ∈ H⊥.

(iii) If u ∈ H and v, w ∈ H⊥, then u× v × w ∈ H.(iv) If u, v, w ∈ H⊥, then u× v × w ∈ H⊥.(v) If u, v, w ∈ H and x ∈ H⊥, then Φ(x, u, v, w) = 0.

(vi) If x, u, v, w is an orthonormal basis of H, then Φ(x, u, v, w) = ±1.(vii) If e ∈ H⊥ has norm one, then H is a coassociative subspace of Ve := e⊥.

(viii) If e ∈ H has norm one, then H ∩ Ve is an associative subspace of Ve.

A 4–dimensional subspace that satisfies these equivalent conditions is called a Cayleysubspace of W . If the vectors u, v, w ∈W are linearly independent, then

H := spanu, v, w, u× v × w

is a Cayley subspace of W .

Proof. We prove that (i) is equivalent to (v). If (i) holds and u, v, w ∈ H, x ∈ H⊥, thenu × v × w ∈ H and, hence, Φ(x, u, v, w) = 〈x, u × v × w〉 = 0. Conversely, if (v) holdsand u, v, w ∈ H, then 〈x, u × v × w〉 = Φ(x, u, v, w) = 0 for every x ∈ H⊥ and henceu× v × w ∈ H.

We prove that (i) is equivalent to (vi). If (i) holds and x, u, v, w is an orthonormal basisof H, then u× v×w is orthogonal to u, v, w and has norm one. Since u× v×w ∈ H, wemust have x = ±u × v × w. Hence Φ(x, u, v, w) = ±|x|2 = ±1. Conversely, assume (vi),let u, v, w ∈ H be orthonormal, and choose x such that x, u, v, w form an orthonormalbasis of H. Then

〈x, u× v × w〉2 = Φ(x, u, v, w)2 = 1 = |x|2|u× v × w|2.

Hence, u × v × w is a real multiple of x and so u × v × w ∈ H. Since the triple crossproduct is alternating, the general case can be reduced to the orthonormal case by scalingand Gram–Schmidt.

That (vi) is equivalent to (vii) follows from Lemma 4.26 and the fact that Φ|Veis the

coassociative calibration on Ve. Likewise, that (vi) is equivalent to (viii) follows fromLemma 4.7 and the fact that ι(e)Φ|Ve is the associative calibration on Ve.

Thus we have proved that (i), (v), (vi), (vii), (viii) are equivalent. The equivalenceof (i), (ii), (iii) for a unit vector u = e ∈ H follows from Lemma 4.26 with V := Ve andH replaced by H⊥, using the fact that v ×e w = −e× v × w is the cross product on Ve.

The equivalence of (iii) and (iv) follows from the equivalence of (i) and (ii) by inter-changing the roles of Λ and Λ⊥. Thus we have proved the equivalence of conditions (i)–(viii). The last assertion of the lemma follows from (i) and equation (6.8). This provesLemma 6.25.

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7. Cayley calibrations

We assume throughout that W is an 8–dimensional real vector space.

Definition 7.1. A 4–form Φ ∈ Λ4W ∗ is called nondegenerate if for every triple u, v, wof linearly independent vectors in W there is a vector x ∈W such that Φ(u, v, w, x) 6= 0.An inner product on W is called compatible with a 4–form Φ if the mapW 3 →W : (u, v, w) 7→ u× v × w defined by

〈x, u× v × w〉 := Φ(x, u, v, w) (7.2)

is a triple cross product. A 4–form Φ ∈ Λ4W ∗ is called a Cayley-form if it admits acompatible inner product.

Example 7.3. The standard Cayley-form on R8 in coordinates x0, x1, . . . , x7 is given by

Φ0 = e0123 − e0145 − e0167 − e0246 + e0257 − e0347 − e0356

+ e4567 − e2367 − e2345 − e1357 + e1346 − e1256 − e1247.

It is compatible with the standard inner product and induces the standard triple crossproduct on R8 (see Example 5.32). Note that Φ0 ∧ Φ0 = 14 vol.

As in Section 3 we shall see that a compatible inner product, if it exists, is uniquelydetermined by Φ. However, in contrast to Section 3, nondegeneracy is, in the presentsetting, not equivalent to the existence of a compatible inner product, but is only anecessary condition. The goal in this section is to give an intrinsic characterization ofCayley-forms. In particular, we shall see that every Cayley-form satisfies the conditionΦ ∧ Φ 6= 0. It seems to be an open question whether or not every nondegenerate 4–formon W has this property; we could not find a counterexample but also did not see how toprove it. We begin by characterizing compatible inner products.

Lemma 7.4. Fix an inner product on W and a 4–form Φ ∈ Λ4W ∗. Then the followingare equivalent:

(i) The inner product is compatible with Φ.(ii) There is a unique orientation on W , with volume form vol ∈ Λ8W ∗, such that, for

all u, v, w ∈W , we have

ι(v)ι(u)Φ ∧ ι(v)ι(u)Φ ∧ Φ = 6|u ∧ v|2vol. (7.5)

(iii) Choose the orientation on W and the volume form vol ∈ Λ8W ∗ as in (ii). Then,for all u, v, w ∈W , we have

ι(v)ι(u)Φ ∧ ι(w)ι(u)Φ ∧ Φ = 6(|u|2〈v, w〉 − 〈v, u〉〈u,w〉

)vol (7.6)

Each of these conditions implies that Φ is nondegenerate and Φ ∧ Φ 6= 0.

Proof. We prove that (i) implies (ii). Assume the inner product is compatible with Φ andlet W 3 → W : (u, v, w) 7→ u × v × w be the triple cross product on W defined by (7.2).Assume u, v ∈W are linearly independent. Then the subspace

Wu,v := w ∈W : 〈u,w〉 = 〈v, w〉 = 0

39


carries a symplectic form ωu,v : Wu,v ×Wu,v → R and a compatible complex structureJu,v : Wu,v →Wu,v given by

ωu,v(x,w) :=Φ(x, u, v, w)

|u ∧ v|, Ju,vw := −u× v × w

|u ∧ v|.

Equation (6.4) asserts that Ju,v is an isometry on Wu,v and equation (6.6) asserts thatJu,v is skew adjoint. Hence, Ju,v is a complex structure on Wu,v and equation (7.2) showsthat, for all x,w ∈Wu,v, we have

ωu,v(x,w) =〈x, u× v × w〉|u ∧ v|

= −〈x, Ju,vw〉.

Thus the inner product ωu,v(·, Ju,v·) on Wu,v is the one inherited from W . It follows that

ωu,v ∧ ωu,v ∧ ωu,v = 6volu,v, (7.7)

where volu,v ∈ Λ6W ∗u,v denotes the volume form on Wu,v with the symplectic orientation.Since the space of linearly independent pairs u, v ∈ W is connected, there is a uniqueorientation on W such that, for every pair u, v of linearly independent vectors in Wand every symplectic basis e1, . . . , e6 of Wu,v, the basis u, v, e1, . . . , e6 of W is positivelyoriented. Let vol ∈ Λ8W ∗ be the volume form of W ∗ for this orientation. Then

volu,v =1

|u ∧ v|ι(v)ι(u)vol|Wu,v

and, hence, equation (7.5) follows from (7.7). This shows that (i) implies (ii). That (ii)implies (iii) follows by using (7.5) with v replaced by v + w.

We prove that (iii) implies (i). Assume there is an orientation on W such that (7.6)holds, and define the map W 3 → W : (u, v, w) 7→ u × v × w by (7.2). That this mapis alternating and satisfies (6.3) is obvious. We prove that it satisfies (6.4). Fix a unitvector e ∈W and denote

Ve := v ∈W : 〈e, v〉 = 0 , φe := ι(e)Φ|Ve , vole := ι(e)vol|Ve .

Then equation (7.6) asserts that

ι(u)φe ∧ ι(v)φe ∧ φe = 6〈u, v〉vole

for every u ∈ Ve. Hence, φe satisfies condition (i) in Lemma 3.4 and therefore is compatiblewith the inner product. This means that the bilinear map Ve×Ve → Ve : (u, v) 7→ u×e vdefined by 〈u×e v, w〉 := φe(u, v, w) is a cross product on Ve. Since

φe(u, v, w) = Φ(w, u, e, v) = 〈u× e× v, w〉,we have u ×e v = u × e × v. This implies |u × e × v| = |u ∧ v| whenever u and v areorthogonal to e and e has norm one. Using Gram–Schmidt and scaling, we deduce thatour map (u, v, w) 7→ u × v × w satisfies (6.4) and, hence, is a triple cross product. Thuswe have proved that (i), (ii), and (iii) are equivalent. Moreover, condition (ii) impliesthat Φ is nondegenerate and (i) implies that Φ ∧ Φ 6= 0, by Remark 6.21. This provesLemma 7.4.

40


We are now in a position to characterize Cayley-forms intrinsically. A 4–form Φ isnondegenerate if and only if the 2–form ι(v)ι(u)Φ ∈ Λ2W ∗ descends to a symplectic formon the quotient W/spanu, v or, equivalently, the 8–form ι(v)ι(u)Φ ∧ ι(v)ι(u)Φ ∧ Φ isnonzero whenever u, v are linearly independent. The question to be addressed is underwhich additional condition we can find an inner product on W that satisfies (7.5).

Theorem 7.8. A 4–form Φ ∈ Λ4W ∗ admits a compatible inner product if and only if itsatisfies the following condition.

(C)

Φ is nondegenerate and, if u, v, w ∈W are linearly independent and

ι(v)ι(u)Φ ∧ ι(w)ι(u)Φ ∧ Φ = ι(u)ι(v)Φ ∧ ι(w)ι(v)Φ ∧ Φ = 0, (7.9)

then, for all x ∈W , we have

ι(w)ι(u)Φ ∧ ι(x)ι(u)Φ ∧ Φ = 0

⇐⇒ ι(w)ι(v)Φ ∧ ι(x)ι(v)Φ ∧ Φ = 0.(7.10)

If this holds, then the compatible inner product is uniquely determined by Φ.

Proof. See page 45.

To understand condition (C) geometrically, assume Φ satisfies (7.6) for some innerproduct on W . Then

ι(v)ι(u)Φ ∧ ι(w)ι(u)Φ ∧ Φ = 0 ⇐⇒ |u|2〈v, w〉 − 〈v, u〉〈u,w〉 = 0.

Hence, if u, v, w are linearly independent, equation (7.9) asserts that w is orthogonalto u and v. Under this assumption both conditions in (7.10) assert that w and x areorthogonal.

Every Cayley-form Φ induces two orientations on W . First, since the 8–formι(v)ι(u)Φ ∧ ι(v)ι(u)Φ ∧ Φ is nonzero for every linearly independent pair u, v ∈ W andthe space of linearly independent pairs in W is connected, there is a unique orientationon W such that ι(v)ι(u)Φ∧ι(v)ι(u)Φ∧Φ > 0 whenever u, v ∈W are linearly independent.The second orientation of W is induced by the 8–form Φ∧Φ. This leads to the followingdefinition.

Definition 7.11. A Cayley-form Φ ∈ Λ4W ∗ is called positive if the 8–forms Φ∧Φ andι(v)ι(u)Φ ∧ ι(v)ι(u)Φ ∧ Φ induce the same orientation whenever u, v ∈ W are linearlyindependent. It is called negative if it is not positive.

Thus Φ is negative if and only if −Φ is positive. Moreover, it follows from Remark 6.21that a Cayley-form Φ ∈ Λ4W ∗ is positive if and only if the associated triple cross productis positive.

Theorem 7.12. If Φ,Ψ ∈ Λ4W ∗ are two positive Cayley-forms, then there is an auto-morphism g ∈ Aut(W ) such that g∗Φ = Ψ.

Proof. See page 46.

41


Lemma 7.13. Let W be a real vector space and g : W 4 → R be a multi-linear mapsatisfying

g(u, v;w, x) = g(w, x;u, v) = −g(v, u;w, x) (7.14)

for all u, v, w, x ∈W and

g(u, v;u, v) > 0 (7.15)

whenever u, v ∈W are linearly independent. Then the matrices

Λu(v, w) :=

(g(u, v;u, v) g(u, v;u,w)g(u,w;u, v) g(u,w;u,w)

)∈ R2×2, and

A(u, v, w) :=

g(v, w; v, w) g(v, w;w, u) g(v, w;u, v)g(w, u; v, w) g(w, u;w, u) g(w, u;u, v)g(u, v; v, w) g(u, v;w, u) g(u, v;u, v)

∈ R3×3

are positive definite whenever u, v, w ∈W are linearly independent. Moreover, the follow-ing are equivalent:

(i) If u, v, w are linearly independent and g(u, v;w, u) = g(v, w;u, v) = 0, then, for allx ∈W , we have

g(u,w;u, x) = 0 ⇐⇒ g(v, w; v, x) = 0. (7.16)

(ii) If u, v, w and u, v, w′ are linearly independent, then

det(Λu(v, w))

det(Λv(u,w))=

det(Λu(v, w′))

det(Λv(u,w′)). (7.17)

(iii) If u, v, w and u, v′, w′ are linearly independent, then

det(Λu(v, w))√det(A(u, v, w))

=det(Λu(v′, w′))√det(A(u, v′, w′))

. (7.18)

(iv) There is an inner product on W such that

g(u, v;u, v) = |u|2|v|2 − 〈u, v〉2 (7.19)

for all u, v ∈W .

If these equivalent conditions are satisfied, then the inner product in (iv) is uniquelydetermined by g and it satisfies

det(Λu(v, w)) = |u|2|u ∧ v ∧ w|2,

det(A(u, v, w)) = |u ∧ v ∧ w|4.(7.20)

42


Proof. Let u, v, w ∈ W be linearly independent. We prove that the matrices Λu(v, w)and A(u, v, w) are positive definite. By (7.15) they have positive diagonal entries. Sincethe determinant of Λu(v, w) agrees with the determinant of the lower right 2× 2 block ofA(u, v, w), it suffices to prove that both matrices have positive determinants. To see this,we observe that the determinants of Λu(v, w) and A(u, v, w) remain unchanged if we addto v a multiple of u and to w a linear combination of u and v. With the appropriate choicesboth matrices become diagonal and thus have positive determinants. Hence, Λu(v, w) andA(u, v, w) are positive definite, as claimed.

We prove that (iv) implies (7.20). The matrix Λu(v, w) and |u ∧ v ∧ w|2 remain un-changed if we add to v and w multiples of u. Hence, we may assume that v and w areorthogonal to u. In this case

Λu(v, w) = |u|2(|v|2 〈v, w〉〈w, v〉 |w|2

)and this implies the first equation in (7.20). Since the determinant of the matrix A(u, v, w)remains unchanged if we add to v a multiple of u and to w a linear combination of u andv, we may assume that u, v, w are pairwise orthogonal. In this case the second equationin (7.20) is obvious. Thus we have proved that (iv) implies (7.20). By (7.20) the innerproduct is uniquely determined by g.

We prove that (i) implies (ii). Fix two linearly independent vectors u, v ∈ W . Thenthe subspace

Wu,v := w ∈W : g(u, v;w, u) = g(v, w;u, v) = 0has codimension two and W = Wu,v ⊕ spanu, v. Now fix an element w ∈ Wu,v.Then (7.16) asserts that the linear functionals x 7→ g(u,w;u, x) and x 7→ g(v, w; v, x)on W have the same kernel. Hence, there exists a constant λ ∈ R such that

g(v, w; v, x) = λg(u,w;u, x)

for all x ∈W . With x = w we obtain λ = g(v, w; v, w)/g(u,w;u,w) and hence

g(u,w;u, x)g(v, w; v, w) = g(u,w;u,w)g(v, w; v, x) for all x ∈W.

This equation asserts that the differential of the map

Wu,v \ 0 → R : w 7→ g(u,w;u,w)

g(v, w; v, w)

vanishes and so the map is constant. This proves (7.17) for all w,w′ ∈ Wu,v \ 0.Since adding to w a linear combination of u and v does not change the determinantsof Λu(v, w) and Λv(u,w), equation (7.17) continues to hold for all w,w′ ∈ W that arelinearly independent of u and v. Thus we have proved that (i) implies (ii).

We prove that (ii) implies (iii). It follows from (7.17) that

w,w′ ∈Wu,v \ 0 =⇒ g(u,w;u,w)

g(v, w; v, w)=g(u,w′;u,w′)

g(v, w′; v, w′).

43


Using this identity with u replaced by u+ v we obtain

w,w′ ∈Wu,v \ 0 =⇒ g(u,w; v, w)

g(v, w; v, w)=g(u,w′; v, w′)

g(v, w′; v, w′).

Now let w,w′ ∈ Wu,v and assume that g(u,w; v, w) = 0. Then we also haveg(u,w′; v, w′) = 0 and so it follows from the definition of Wu,v that all off-diagonal termsin the matrices Λu(v, w), Λu(v, w′), A(u, v, w), and A(u, v, w′) vanish. Hence,

det(Λu(v, w))2

det(A(u, v, w))=g(u, v;u, v)g(u,w;u,w)

g(v, w; v, w)

=g(u, v;u, v)g(u,w′;u,w′)

g(v, w′; v, w′)=

det(Λu(v, w′))2

det(A(u, v, w′)).

Thus we have proved (7.18) under the assumption that w,w′ ∈ Wu,v \ 0 andg(w, u;w, v) = 0. Since the determinants of Λu(v, w) and A(u, v, w) remain unchangedif we add to w a linear combination of u and v and if we add to v a multiple of u,equation (7.18) continues to hold when v = v′. If u, v, w and u, v, w′ and u, v′, w′ are alllinearly independent triples, we obtain

det(Λu(v, w))2

det(A(u, v, w))=

det(Λu(v, w′))2

det(A(u, v, w′))=

det(Λu(v′, w′))2

det(A(u, v′, w′)).

Here the last equation follows from the first by symmetry in v and w. This provesequation (7.18) under the additional assumption that u, v, w′ is a linearly independenttriple. This assumption can be dropped by continuity. Thus we have proved that (ii)implies (iii).

We prove that (iii) implies (iv). Define a function W → [0,∞) : u 7→ |u| by |u| := 0for u = 0 and by

|u|2 :=g(u,w;u,w)g(u, v;u, v)− g(u, v;u,w)2√

det(A(u, v, w))(7.21)

for u 6= 0, where v, w ∈W are chosen such that u, v, w are linearly independent. By (7.18)the right hand side of (7.21) is independent of v and w. It follows from (7.21) with ureplaced by u+ v that

|u+ v|2 − |u|2 − |v|2 = 2g(u,w; v, w)g(u, v;u, v)− g(u, v;u,w)g(u, v; v, w)√

det(A(u, v, w)).

Replacing v by −v gives |u+ v|2 + |u− v|2 = 2|u|2 + 2|v|2. Thus the map

W → [0,∞) : u 7→ |u|is continuous, satisfies the parallelogram identity, and vanishes only for u = 0. Hence,it is a norm on W and the associated inner product of two linearly independent vectorsu, v ∈W is given by

〈u, v〉 :=g(u,w; v, w)g(u, v;u, v)− g(u, v;u,w)g(u, v; v, w)√

det(A(u, v, w))(7.22)

44


whenever w ∈ W is chosen such that u, v, w are linearly independent. That this innerproduct satisfies (7.19) for every pair of linearly independent vectors follows from (7.21)and (7.22) with w ∈Wu,v. This proves that (iii) implies (iv).

We prove that (iv) implies (i). Replacing v in equation (7.19) by v + w we obtain

g(u, v;u,w) = |u|2〈v, w〉 − 〈u, v〉〈u,w〉.

for all u, v, w ∈W . Hence,

g(u, v;w, u) = g(v, w;u, v) = 0 ⇐⇒ 〈u,w〉 = 〈v, w〉 = 0.

If w ∈W is orthogonal to u and v, then we have

g(u,w;u, x) = |u|2〈w, x〉

and

g(v, w; v, x) = |v|2〈w, x〉.This implies (7.16) and proves Lemma 7.13.

Proof of Theorem 7.8. If Φ is nondegenerate and u ∈W is nonzero, then ι(u)Φ descendsto a nondegenerate 3–form on the 7–dimensional quotient space W/Ru. By Lemma 3.4this implies that ι(v)ι(u)Φ∧ ι(v)ι(u)Φ∧ ι(u)Φ descends to a nonzero 7–form on W/Ru forevery vector v ∈ W \Ru. Hence, the 8–form ι(v)ι(u)Φ ∧ ι(v)ι(u)Φ ∧ Φ on W is nonzerowhenever u, v are linearly independent. The orientation on W induced by this form isindependent of the choice of the pair u, v. Choose any volume form Ω ∈ Λ8W ∗ compatiblewith this orientation and, for λ > 0, define a multi-linear function gλ : W 4 → R by

gλ(u, v;w, x) :=ι(v)ι(u)Φ ∧ ι(x)ι(w)Φ ∧ Φ

6λ4Ω(7.23)

This function satisfies (7.14) and (7.15) and, if Φ satisfies (C), it also satisfies (7.16).Hence, it follows from Lemma 7.13 that there is a unique inner product 〈·, ·〉λ on W suchthat, for all u, v ∈W , we have

gλ(u, v;u, v) = |u|2λ|v|2λ − 〈u, v〉

2λ. (7.24)

Let volλ be the volume form associated to the inner product and the orientation. Thenthere is a constant µ(λ) > 0 such that

volλ = µ(λ)2Ω.

We have gλ = λ−4g1, hence |u|λ = λ−1|u|1 for every u ∈ W , and hence volλ = λ−8vol1.

Thus µ(λ) = λ−4µ(1). With λ := µ(1)1/6 we obtain µ(λ) = λ−4µ(1) = µ(1)1/3 = λ2.With this value of λ we have λ4Ω = volλ. Hence, it follows from (7.23) and (7.24) that

ι(v)ι(u)Φ ∧ ι(v)ι(u)Φ ∧ Φ = 6(|u|2λ|v|

2λ − 〈u, v〉

2λ

)volλ.

Hence, by Lemma 7.4, Φ is compatible with the inner product 〈·, ·〉λ. This shows thatevery 4–form Φ ∈ Λ4W ∗ that satisfies (C) is compatible with a unique inner product.

45


Conversely, suppose that Φ is compatible with an inner product. Then, by Lemma 7.4,there is an orientation on W such that the associated volume form vol ∈ Λ8W ∗ satis-fies (7.5). Define g : W 4 → R by

g(u, v;w, x) :=ι(v)ι(u)Φ ∧ ι(x)ι(w)Φ ∧ Φ

6vol.

By (7.5) this map satisfies condition (iv) in Lemma 7.13 and it obviously satisfies (7.14)and (7.15). Hence, it satisfies condition (i) in Lemma 7.13 and this implies that Φ satis-fies (C). This proves Theorem 7.8.

Proof of Theorem 7.12. Let Φ ∈ Λ4W ∗ be a positive Cayley-form with the associatedinner product, orientation, and triple cross product. Let φ0 ∈ Λ3(R7)∗ and ψ0 ∈ Λ4(R7)∗

be the standard associative and coassociative calibrations defined in Example 2.15 and inthe proof of Lemma 4.8. Then Φ0 := 1∗∧φ0 +ψ0 ∈ Λ4(R8)∗ is the standard Cayley-formon R8.

Choose a unit vector e ∈W and denote

Ve := e⊥, φe := ι(e)Φ|Ve∈ Λ3V ∗e , ψe := Φ|Ve

∈ Λ4V ∗e .

Then φe is a nondegenerate 3–form on Ve and, hence, by Theorem 3.2, there is an iso-morphism g : R7 → Ve such that g∗φe = φ0. It follows also from Theorem 3.2 that gidentifies the standard inner product on R7 with the unique inner product on Ve that iscompatible with φe, and the standard orientation on R7 with the orientation determinedby φe via Lemma 3.4. Hence, it follows from Lemma 4.8 that g also identifies the twocoassociative calibrations, i.e., g∗ψe = ψ0. Since Φ is a positive Cayley-form, we have

Φ = e∗ ∧ φe + ψe.

Hence, if we extend g to an isomorphism R8 = R⊕R7 →W , which is still denoted by gand sends e0 = 1 ∈ R ⊂ R8 to e, we obtain g∗Φ = Φ0 and this proves Theorem 7.12.

Remark 7.25. The space S2Λ2W ∗ of symmetric bilinear forms on Λ2W can be identi-fied with the space of multi-linear maps g : W 4 → R that satisfy (7.14). Denote byS2

0Λ2W ∗ ⊂ S2Λ2W ∗ the subspace of all g ∈ S2Λ2W ∗ that satisfy the algebraic Bianchiidentity

g(u, v;w, x) + g(v, w;u, x) + g(w, u; v, x) = 0 (7.26)

for all u, v, w, x ∈W . Then there is a direct sum decomposition

S2Λ2W ∗ = Λ4W ∗ ⊕ S20Λ2W ∗

and the projection

Π : S2Λ2W ∗ → Λ4W ∗

is given by

(Πg)(u, v, w, x) := 13

(g(u, v;w, x) + g(v, w;u, x) + g(w, u; v, x)

).

46


Note that

dim Λ2W = 28, dim S2Λ2W = 406, (7.27)

dim Λ4W = 70, dim S20Λ2W = 336. (7.28)

Moreover, there is a natural quadratic map qΛ : S2W ∗ → S20Λ2W ∗ given by(

qΛ(γ))(u, v;x, y) := γ(u, x)γ(v, y)− γ(u, y)γ(v, x)

for γ ∈ S2W ∗ and u, v, x, y ∈ W . Lemma 7.13 asserts, in particular, that the restrictionof this map to the subset of inner products is injective and, for each element g ∈ S2Λ2W ∗,it gives a necessary and sufficient condition for the existence of an inner product γ on Wsuch that

g −Πg = qΛ(γ).

We shall see in Corollary 9.9 below that, if Φ ∈ Λ4W ∗ is a positive Cayley-form andg = gΦ ∈ S2Λ2W ∗ is given by

gΦ(u, v;x, y) :=ι(v)ι(u)Φ ∧ ι(y)ι(x)Φ ∧ Φ

vol, vol :=

Φ ∧ Φ

14,

then

gΦ = 6qΛ(γ) + 7Φ

for a unique inner product γ ∈ S2W ∗, and the volume form of γ is indeed vol. Thus, inparticular, we have ΠgΦ = 7Φ.

Remark 7.29. The space S2S2W ∗ of symmetric bilinear forms on S2W can be identifiedwith the space of multi-linear maps σ : W 4 → R that satisfy

σ(u, v;x, y) = σ(x, y;u, v) = σ(v, u;x, y). (7.30)

Denote by S20S

2W ∗ the subspace of all σ ∈ S2S2W ∗ that satisfy the algebraic Bianchiidentity (7.26). Then

S2S2W ∗ = S4W ∗ ⊕ S20S

2W ∗,

where

dim S2W = 36, dim S2S2W = 666, (7.31)

dim S4W = 330, dim S20S

2W = 336. (7.32)

The projection Π : S2S2W ∗ → S4W ∗ is given by the same formula as above. Thus

(σ −Πσ)(u, v;x, y) = 23σ(u, v;x, y)− 1

3σ(v, x;u, y)− 13σ(x, u; v, y).

There is a natural quadratic map qS : S2W ∗ → S2S2W ∗ given by(qS(γ)

)(u, v;x, y) := γ(u, v)γ(x, y).

Polarizing the quadratic map qΛ : S2W ∗ → S2Λ2W ∗ one obtains a linear map

T : S2S2W ∗ → S2Λ2W ∗

47


given by (Tσ)(u, v;x, y) := σ(u, x; v, y)− σ(u, y; v, x)

such that qΛ = T qS . The image of T is the subspace S20Λ2W ∗ of solutions of the

algebraic Bianchi identity (7.26) and its kernel is the subspace S4W ∗. A pseudo-inverseof T is the map S : S2Λ2W ∗ → S2S2W ∗ given by(

Sg)(u, v;x, y) := 1

3

(g(u, x; v, y) + g(u, y; v, x)

)whose kernel is Λ4W ∗ and whose image is S2

0S2W ∗. Thus

TSg = g −Πg, STσ = σ −Πσ

for g ∈ S2Λ2W ∗ and σ ∈ S2S2W ∗. Given g ∈ S2Λ2W ∗ and γ ∈ S2W ∗, we have

g −Πg = qΛ(γ) ⇐⇒ Sg = (1−Π)qS(γ).

Namely, if qΛ(γ) = g − Πg, then Sg = S(g − Πg) = SqΛ(γ) = qS(γ) − ΠqS(γ), and if(1−Π)qS(γ) = Sg, then (1−Π)g = TSg = T (1−Π)qS(γ) = qΛ(γ).

8. The group G2

Let V be a 7–dimensional real Hilbert space equipped with a cross product and letφ ∈ Λ3V ∗ be the associative calibration defined by (2.8). We orient V as in Lemma 3.4and denote by

∗ : ΛkV ∗ → Λ7−kV ∗

the associated Hodge ∗–operator and by ψ := ∗φ ∈ Λ4V ∗ the coassociative calibration.Recall that V is equipped with an associator bracket via (4.1), related to ψ via (4.9), andwith a coassociator bracket (4.22).

The group of automorphisms of φ will be denoted by

G(V, φ) := g ∈ GL(V ) : g∗φ = φ .

By Lemma 2.20, we have G(V, φ) ⊂ SO(V ) and hence, by (2.8),

G(V, φ) = g ∈ SO(V ) : gu× gv = g(u× v) ∀u, v ∈ V .

For the standard structure φ0 on R7 in Example 2.15 we denote the structure group byG2 := G(R7, φ0). By Theorem 3.2, the group G(V, φ) is isomorphic to G2 for everynondegenerate 3–form on a 7–dimensional vector space.

Theorem 8.1. The group G(V, φ) is a 14–dimensional simple, connected, simply con-nected Lie group. It acts transitively on the unit sphere and, for every unit vector u ∈ V ,the isotropy subgroup Gu := g ∈ G(V, φ) : gu = u is isomorphic to SU(3). Thus thereis a fibration

SU(3) → G2 −→ S6.

Proof. As we have observed in Step 4 in the proof of Lemma 3.4, the group G = G(V, φ)has dimension at least 14, as it is an isotropy subgroup of the action of the 49–dimensional

48


group GL(V ) on the 35–dimensional space Λ3V ∗. Since G ⊂ SO(V ), by Lemma 2.20, thegroup acts on the unit sphere

S := u ∈ V : |u| = 1 .

Thus, for every u ∈ S, the isotropy subgroup Gu has dimension at least 8. By Lemma 2.18,the group Gu preserves the subspace Wu := u⊥, the symplectic form ωu, and the complexstructure Ju on Wu given by ωu(v, w) = 〈u, v × w〉 and Juv = u × v. Hence, Gu isisomorphic to a subgroup of U(Wu, ωu, Ju) ∼= U(3). Now consider the complex valued3–form θu ∈ Λ3,0W ∗u given by

θu(x, y, z) := φ(x, y, z)− iφ(u× x, y, z) = φ(x, y, z)− iψ(u, x, y, z)

for x, y, z ∈ Wu. (See (4.1) and (4.9) for the last equality.) This form is nonzero andis preserved by Gu. Hence, Gu is isomorphic to a subgroup of SU(Wu, ωu, Ju). SinceSU(Wu, ωu, Ju) ∼= SU(3) is a connected Lie group of dimension 8 and Gu has dimensionat least 8, it follows that

Gu∼= SU(Wu, ωu, Ju) ∼= SU(3).

In particular, dim Gu = 8 and so dim G ≤ dim Gu+dim S = 14. This implies dim G = 14and, since S is connected, G acts transitively on S. Thus we have proved that there is afibration SU(3) → G→ S. It follows from the homotopy exact sequence of this fibrationthat G is connected and simply connected and that π3(G) ∼= Z. Hence, G is simple.

Here is another proof that G is simple. Let g := Lie(G) denote its Lie algebra and, forevery u ∈ S, let gu := Lie(Gu) denote the Lie algebra of the isotropy subgroup. Then,for every ξ ∈ g, we have ξ ∈ gu if and only if u ∈ ker ξ. Since every ξ ∈ g is skew-adjoint,it has a nontrivial kernel and hence belongs to gu for some u ∈ S.

Now let I ⊂ g be a nonzero ideal. Then, by what we have just observed, there is anelement u ∈ S such that I ∩ gu 6= 0. Thus I ∩ gu is a nonzero ideal in gu and, sincegu is simple, this implies gu ⊂ I. Next we claim that, for every v ∈ u⊥, there is anelement ξ ∈ I such that ξu = v. To see this, choose any element η ∈ gu ⊂ I such thatker η = 〈u〉. Then there is a unique element w ∈ u⊥ such that ηw = v. Since G actstransitively on S there is an element ζ ∈ g such that ζu = w. Hence, ξ = [η, ζ] ∈ I andξu = ηζu = ηw = v. This proves that dim(I/gu) ≥ 6; hence, dim I ≥ 14, and henceI = g. This proves Theorem 8.1.

We examine the action of the group G(V, φ) on the space

S :=

(u, v, w) ∈ V :

|u| = |v| = |w| = 1,〈u, v〉 = 〈u,w〉 = 〈v, w〉 = 〈u× v, w〉 = 0

.

Let S ⊂ V denote the unit sphere. Then each tangent space TuS = u⊥ carries a naturalcomplex structure v 7→ u × v. The space S is a bundle over S whose fiber over u isthe space of Hermitian orthonormal pairs in TuS. Hence, S is a bundle of 3–spheresover a bundle of 5–spheres over a 6–sphere and therefore is a compact connected simplyconnected 14–dimensional manifold.

49


Theorem 8.2. The group G(V, φ) acts freely and transitively on S .

Proof. We give two proofs of this result. The first proof uses the fact that the isotropysubgroup Gu ⊂ G := G(V, φ) of a unit vector u ∈ V is isomorphic to SU(3) and theisotropy subgroup in SU(3) of a Hermitian orthonormal pair is the identity. Hence, Gacts freely on S . Since G and S are compact connected manifolds of the same dimension,this implies that G acts transitively on S .

For the second proof we assume that φ = φ0 is the standard structure on V = R7.Given (u, v, w) ∈ S , define g : R7 → R7 by

ge1 = u, ge2 = v, ge3 = u× v, ge4 = w

ge5 = w × u, ge6 = w × v, ge7 = w × (u× v).

By construction g preserves the cross product and the inner product. Hence, g ∈ G2.Moreover, g is the unique element of G2 that maps the triple (e1, e2, e4) to (u, v, w). Thisproves Theorem 8.2.

Corollary 8.3. The group G(V, φ) acts transitively on the space of associative subspacesof V and on the space of coassociative subspaces of V .

Proof. This follows from Theorem 8.2, Lemma 4.7, and Lemma 4.26.

Remark 8.4. Let Λ ⊂ V be an associative subspace and define H := Λ⊥ and

GΛ := g ∈ G(V, φ) : gΛ = Λ.Then every h ∈ SO(H) extends uniquely to an element g ∈ GΛ (choose (u, v, w) ∈ S suchthat u, v, w ∈ H) and the action of g on Λ is induced by the action of h on Λ+H∗ under theisomorphism in Remark 4.27. Hence the map GΛ → SO(H) : g 7→ g|H is an isomorphismand so the associative Grassmannian L := Λ ⊂ V : Λ is an associative subspace isdiffeomorphic to the homogeneous space G(V, φ)/SO(H) ∼= G2/SO(4), by Corollary 8.3.Since Λ ⊂ V is associative if and only if H := Λ⊥ is coassociative (see Lemma 4.26), Lalso is the coassociative Grassmannian.

Theorem 8.5. There are orthogonal splittings

Λ2V ∗ = Λ27 ⊕ Λ2

14,

Λ3V ∗ = Λ31 ⊕ Λ3

7 ⊕ Λ327,

where dim Λkd = d and

Λ27 := ι(u)φ : u ∈ V =

ω ∈ Λ2V ∗ : ∗(φ ∧ ω) = 2ω

,

Λ214 :=

ω ∈ Λ2V ∗ : ψ ∧ ω = 0

=ω ∈ Λ2V ∗ : ∗(φ ∧ ω) = −ω

,

Λ31 := 〈φ〉,

Λ37 := ι(u)ψ : u ∈ V ,

Λ327 :=

ω ∈ Λ3V ∗ : φ ∧ ω = 0, ψ ∧ ω = 0

.

50


Each of the spaces Λkd is an irreducible representation of G(V, φ) and the representationsΛ2

7 and Λ37 are both isomorphic to V , Λ2

14 is isomorphic to the Lie algebrag(V, φ) := Lie(G(V, φ)) ∼= g2, and Λ3

27 is isomorphic to the space of traceless symmetricendomorphisms of V . The orthogonal projections π7 : Λ2V ∗ → Λ2

7 and π14 : Λ2V ∗ → Λ214

are given by

π7(ω) = 13ω + 1

3 ∗ (φ ∧ ω) = 13 ∗(ψ ∧ ∗(ψ ∧ ω)

), (8.6)

π14(ω) = 23ω −

13 ∗ (φ ∧ ω) = ω − 1

3 ∗(ψ ∧ ∗(ψ ∧ ω)

). (8.7)

Proof. For u ∈ V denote by Au ∈ so(V ) the endomorphism Auv := u× v. Then the Liealgebra g := Lie(G) of G = G(V, φ) is given by

g = ξ ∈ End(V ) : ξ + ξ∗ = 0, Aξu + [Au, ξ] = 0 ∀u ∈ V .

Step 1. There is an orthogonal decomposition

so(V ) = g⊕ h, h := Au : u ∈ V with respect to the inner product 〈ξ, η〉 := − 1

2 tr(ξη) on so(V ).

The group G acts on the space so(V ) of skew-adjoint endomorphisms by conjugationand this action preserves the inner product. Both subspaces g and h are invariant underthis action, because gAug

−1 = Agu for all u ∈ V and g ∈ G. If ξ = Au ∈ g ∩ h, then0 = LAu

φ = 3ι(u)ψ (see equation (4.19)) and hence u = 0. This shows that g ∩ h = 0.Since dim g = 14, dim h = 7, and dim so(V ) = 21, we have so(V ) = g⊕ h. Moreover, g⊥

is another G–invariant complement of g. Hence h is the graph of a G–equivariant linearmap g⊥ → g. The image of this map is an ideal in g and hence must be zero. This showsthat h = g⊥.

Step 2. Λ214 is the orthogonal complement of Λ2

7

By equation (4.41) in Lemma 4.37 we have u∗ ∧ ψ = ∗ι(u)φ for all u ∈ V . Hence,u∗ ∧ ω ∧ ψ = ω ∧ ∗ι(u)φ and this proves Step 2.

Step 3. The isomorphism so(V ) → Λ2V ∗ : ξ 7→ ωξ := 〈·, ξ·〉 is an SO(V )–equivariantisometry and maps g onto Λ2

14

That the isomorphism ξ 7→ ωξ is an SO(V )–equivariant isometry follows directly fromthe definitions. The image of h under this isomorphism is obviously the subspace Λ2

7.Hence, by Step 1, the orthogonal complement of Λ2

7 is the image of g under this isomor-phism. Hence, the assertion follows from Step 2.

Step 4. Let ω ∈ Λ2V ∗. Then ψ ∧ ω = 0 if and only if ∗(φ ∧ ω) = −ω.

Define the operators Q : Λ2V ∗ → Λ2V ∗ and R : Λ2V ∗ → Λ1V ∗ by

Qω := ∗(φ ∧ ω), Rω := ∗(ψ ∧ ω)

for ω ∈ Λ2V ∗. Then Q is self-adjoint and R∗ : Λ1V ∗ → Λ2V ∗ is given by the sameformula R∗α = ∗(ψ ∧ α) for α ∈ Λ1V ∗. Both operators are G–equivariant. Moreover,

51


R∗R = Q+id by equation (4.60) in Lemma 4.37. Hence, Rω = 0 if and only if Qω = −ω.(Note also that the operator R∗R vanishes on Λ2

14 by equation (4.60) and has eigenvalue3 on Λ2

7 by (4.52).) This proves Step 4.One can rephrase this argument more geometrically as follows. The action of G on Λ2

14

is irreducible by Step 3. Hence, Λ214 is (contained in) an eigenspace of the operator Q.

Moreover, the operator Q is traceless. To see this, let e1, . . . , e7 be an orthonormal basisof V and denote by e1, . . . , e7 the dual basis of V ∗. Then the 2–forms eij := ei ∧ ej withi < j form an orthonormal basis of Λ2V ∗ and we have∑

i<j

〈eij , ∗(φ ∧ eij)〉 =∑i<j

(eij ∧ eij ∧ φ)(e1, . . . , e7) = 0.

By equation (4.49) in Lemma 4.37, the operator Q has eigenvalue 2 on the 7–dimensionalsubspace Λ2

7. Since dim Λ2V ∗ = 21, it follows that Q has eigenvalue −1 on the 14–dimensional subspace Λ2

14. This gives rise to another proof of equation (4.60) and com-pletes the second proof of Step 4.

Step 5. The subspaces Λ31, Λ3

7, and Λ327 form an orthogonal decomposition of Λ3V ∗ and

dim Λ3d = d.

That dim Λ3d = d for d = 1, 7 is obvious. Since ∗ι(u)ψ = −u∗ ∧ φ, it follows that Λ3

1 isorthogonal to Λ3

7. Moreover, for every ω ∈ Λ3V ∗, we have

φ ∧ ω = 0 ⇐⇒ u∗ ∧ φ ∧ ω = 0 ∀u ∈ V ⇐⇒ ω ⊥ Λ37

and

ψ ∧ ω = 0 ⇐⇒ ω ⊥ Λ31.

Hence, Λ327 is the orthogonal complement of Λ3

1 ⊕ Λ37. Since dim Λ3V ∗ = 35, this proves

Step 5.

Step 6. The subspaces Λ27, Λ2

14, Λ31, Λ3

7, Λ327 are irreducible representations of the group

G = G(V, φ).

The irreducibility of Λ31 and Λ2

7∼= Λ3

7 is obvious and for Λ214 it follows from Step 3. We

also point out that Λ37 is the tangent space of the orbit of φ under the action of SO(V ).

The space Λ327 can be identified with the space of traceless symmetric endomorphisms

S : V → V via S 7→ LSφ by Theorem 8.8 below. That it is an irreducible representationof G(V, φ) is shown in [Bry87]. This proves Step 6. Equations (8.6) and (8.7) followdirectly from the definitions and (4.60). This proves Theorem 8.5.

Theorem 8.8. The linear map

End(V )→ Λ3V ∗ : A 7→ LAφ

(see Remark 4.16) restricts to a G(V, φ)–equivariant isomorphism from the space of trace-less symmetric endomorphisms of V onto Λ3

27.

52


Proof. We follow the exposition of Karigiannis in [Kar09, Section 2]. Define the linearmap Λ3V ∗ → End(V ) : η 7→ Sη by

〈u, Sηv〉 :=ι(u)φ ∧ ι(v)φ ∧ η

4vol(8.9)

for η ∈ Λ3V ∗ and u, v ∈ V . This map has the following properties.

Step 1. Let A ∈ End(V ). Then

SLAφ = 12 (A∗ +A) + 1

2 tr(A)1. (8.10)

In particular, Sφ = 321.

For t ∈ R define gt := eAt and φt := g∗t φ. Then φt ∈ Λ3V ∗ is a nondegenerate 3–formcompatible with the inner product

〈u, v〉t := 〈gtu, gtv〉

on V and the volume form volt ∈ Λ7V ∗ given by

volt := g∗t vol = det(gt)vol.

Hence,

ι(u)φt ∧ ι(u)φt ∧ φt = 6|u|2tvolt

for all u ∈ V and all t ∈ R. Differentiate this equation with respect to t at t = 0 and usethe identity 0 = ι(u)(ι(u)φ ∧ φ ∧ η) = ι(u)φ ∧ ι(u)φ ∧ η − ι(u)φ ∧ φ ∧ ι(u)η for η ∈ Λ3V ∗

to obtain

3ι(u)φ ∧ ι(u)φ ∧ LAφ = 12〈u,Au〉vol + 6|u|2 tr(A)vol.

Divide this equation by 12vol and use the definition of SLAφ in equation (8.9) to obtain

〈u, SLAφu〉 = 〈u,Au〉+ 12 tr(A)|u|2.

Since SLAφ is a symmetric endomorphism, this proves equation (8.10). Now take A = 1and use the identities L1φ = 3φ and tr(1) = 7 to obtain S3φ = SL1φ = 9

21. This provesStep 1.

Step 2. Let v ∈ V . Then Sι(v)ψ = 0.

It follows from equation (4.47) in Lemma 4.37 that

ι(u)φ ∧ α ∧ ψvol

=α ∧ ∗u∗

vol= 3α(u) (8.11)

for all u ∈ V and all α ∈ V ∗. Take α := ι(w)ι(v)φ = φ(v, w, ·) to obtain

3φ(u, v, w) =ι(u)φ ∧ ι(w)ι(v)φ ∧ ψ

vol. (8.12)

Interchange u and v to obtain

− 3φ(u, v, w) =ι(w)ι(u)φ ∧ ι(v)φ ∧ ψ

vol. (8.13)

53


Now contract the vector w with the 8–form ι(u)φ ∧ ι(v)φ ∧ ψ = 0 to obtain

0 = ι(w)(ι(u)φ ∧ ι(v)φ ∧ ψ

)= ι(w)ι(u)φ ∧ ι(v)φ ∧ ψ

+ ι(u)φ ∧ ι(w)ι(v)φ ∧ ψ+ ι(u)φ ∧ ι(v)φ ∧ ι(w)ψ

= ι(u)φ ∧ ι(v)φ ∧ ι(w)ψ.

Here the last step follows from (8.12) and (8.13). Thus we have proved that

ι(u)φ ∧ ι(v)φ ∧ ι(w)ψ = 0 for all u, v, w ∈ V. (8.14)

Hence, Sι(w)ψ = 0 for all w ∈ V by definition of Sη. This proves Step 2.

Step 3. Let S = S∗ ∈ End(V ) be a self-adjoint endomorphism. Then

∗ LSφ = tr(S)ψ − LSψ. (8.15)

It suffices to prove this for self-adjoint rank 1 endomorphisms. Let u ∈ V and defineS := uu∗. Then tr(S) = |u|2 and LSφ = u∗ ∧ ι(u)φ. Hence,

∗LSφ = ∗(u∗ ∧ ι(u)φ)

= ∗(u∗ ∧ ∗(u∗ ∧ ψ)

)= ι(u)(u∗ ∧ ψ)

= |u|2ψ − u∗ ∧ ι(u)ψ

= tr(S)ψ − LSψ.

Here the third step uses the identity u∗∧∗α = (−1)k−1 ∗ ι(u)α in Remark 4.14 with k = 5and α = u∗ ∧ ψ. This proves Step 3.

Step 4. Let S = S∗ ∈ End(V ) and T = T ∗ ∈ End(V ) be self-adjoint endomorphisms.Then

〈LSφ,LTφ〉 = 2 tr(ST ) + tr(S) tr(T ). (8.16)

It suffices to prove this for self-adjoint rank 1 endomorphisms. Let u, v ∈ V anddefine S := uu∗ and T := vv∗. Then tr(S) = |u|2, tr(T ) = |v|2, tr(ST ) = 〈u, v〉2,LSφ = u∗ ∧ ι(u)φ, LTφ = v∗ ∧ ι(v)φ. Hence, by Step 3,

〈LSφ,LTφ〉vol = LSφ ∧ ∗LTφ= LSφ ∧

(tr(T )ψ − LTψ

)= |v|2u∗ ∧ ι(u)φ ∧ ψ − u∗ ∧ ι(u)φ ∧ v∗ ∧ ι(v)ψ

= |v|2ι(u)φ ∧ ∗ι(u)φ− u∗ ∧ v∗ ∧ ι(u)φ ∧ ι(v)ψ

=(3|u|2|v|2 − 2|u× v|2

)vol

=(|u|2|v|2 + 2〈u, v〉2

)vol.

54


Here the fourth step follows from (4.41) and the fifth step follows from (4.43) and (4.56).This proves Step 4.

Step 5. Let S = S∗ ∈ End(V ) be a self-adjoint endomorphism and let u ∈ V . Then〈ι(u)ψ,LSφ〉 = 0.

It suffices to prove this for rank 1 endomorphisms. Let v ∈ V and define S := vv∗.Then ∗LSφ = tr(S)ψ − LSψ = |v|2ψ − v∗ ∧ ι(v)ψ by Step 3, so

ι(u)ψ ∧ ∗LSφ = |v|2ι(u)ψ ∧ ψ − ι(u)ψ ∧ v∗ ∧ ι(v)ψ = 0.

Here the last equation follows from (4.40) and (4.48).

Step 6. Define

Endsym0 (V ) := S ∈ End(V ) : S = S∗, tr(S) = 0 .

Then the map A 7→ LAφ restricts to G(V, φ)–equivariant isomorphism

Endsym0 (V )→ Λ3

27 : S 7→ LSφ.

That the map A 7→ LAφ is G(V, φ)–equivariant follows directly from the definitions.Now let S ∈ Endsym

0 (V ). Then by Step 4

LSφ ∧ ψvol

= 〈LSφ, φ〉 = 13 〈LSφ,L1φ〉 = tr(S) = 0.

Moreover, ∗ι(u)ψ = −u∗ ∧ φ by (4.42) and so u∗ ∧ LSφ ∧ φ = −〈LSφ, ι(u)ψ〉 = 0 for allu ∈ V by Step 5. This shows that LSφ ∧ φ = 0 and LSφ ∧ ψ = 0, and so LSφ ∈ Λ3

27.Moreover, SLSφ = S for all S ∈ Endsym

0 (V ) by Step 1. Thus the map Endsym0 (V )→ Λ3

27

given by S 7→ LSφ is injective. Since Endsym0 (V ) and Λ3

27 both have dimension 27, thisproves Step 6 and Theorem 8.8.

The above proof of Theorem 8.8 does not use the fact that the G(V, φ)–representationEndsym

0 (V ), and hence also Λ327, is irreducible. Moreover, we have not included a proof of

this fact in these notes (although it is stated in Theorem 8.5). Assuming irreducibility,the proof of Theorem 8.8 can be simplified as follows.

Proof of Theorem 8.8 assuming Endsym0 (V ) is irreducible. Since

LAφ =d

dt

∣∣∣∣t=0

exp(tA)∗φ,

it is clear that the map End(V ) → Λ3V ∗ : A 7→ LAφ is G(V, φ)–equivariant. Its kernelis Lie(G(V, φ)) and hence its restriction to Endsym

0 (V ) is injective. Now the compositionof the map Endsym

0 (V ) → Λ3V ∗ : A → LAφ with the orthogonal projection onto Λ31,

respectively Λ37, is G(V, φ)–equivariant by Step 5 in the proof of Theorem 8.5. This

composition cannot be an isomorphism for dimensional reasons, and hence must vanishby Schur’s Lemma, because the G(V, φ)–representations Endsym

0 (V ), Λ31, and Λ3

7 are allirreducible. Thus the image of Endsym

0 (V ) under the map A 7→ LAφ is perpendicular toΛ3

1 and Λ37, and hence is equal to Λ3

27.

55


We close this section with the proof of a well-known formula for the differential ofthe map that assigns to a nondegenerate 3–form its coassociative calibration. Let V bea seven-dimensional real vector space, abbreviate Λk := ΛkV ∗ for k = 0, 1, . . . , 7, anddefine

P = P(V ) :=φ ∈ Λ3

∣∣φ is nondegenerate.

This is an open subset of Λ3 and it is diffeomorphic to the homogeneous space GL(7,R)/G2.Namely, if φ0 ∈ P is any nondegenerate 3–form then the map GL(V )→ P : g 7→ (g−1)∗φ0

descends to a diffeomorphism from the quotient space GL(V )/G(V, φ0) to P. Define themap Θ : P → Λ4 by

Θ(φ) := ∗φφ. (8.17)

Here ∗φ : Λ3 → Λ4 denotes the Hodge ∗–operator associated to the inner product andorientation determined by φ.

Theorem 8.18. The map Θ : P → Λ4 in (8.17) is a GL(V )–equivariant local diffeomor-phism, it restricts to a diffeomorphism onto its image on each connected component of P,and its derivative at φ ∈ P is given by

dΘ(φ)η = ∗φ(

43π1(η) + π7(η)− π27(η)

)(8.19)

for η ∈ Λ3. Here πd : Λ3 → Λ3d denotes the projection associated to the orthogonal splitting

Λ3 = Λ31 ⊕ Λ3

7 ⊕ Λ327 in Theorem 8.5 determined by φ.

Proof. That P has two connected components distinguished by the orientation of V fol-lows from the fact that GL(V ) has two connected components. That the restriction of Θto each connected component of P is bijective follows from Theorem 4.30 and that it is adiffeomorphism then follows from equation (8.19) and the inverse function theorem.

Thus it remains to prove (8.19). Since Θ is GL(V )–equivariant, it satisfies

Θ(g∗φ) = g∗Θ(φ) (8.20)

for φ ∈ P and g ∈ GL(V ). Fix a nondegenerate 3–form φ ∈ P, denote by ψ := Θ(φ) = ∗φφits coassociative calibration, and differentiate equation (8.20) at g = 1 in the directionA ∈ End(V ) to obtain

dΘ(φ)LAφ = LAψ. (8.21)

Now let η ∈ Λ3 and denote ηd := πd(η) for d = 1, 7, 27. By Theorem 8.5 and Theorem 8.8there exists a real number λ, a vector u ∈ V , and a traceless symmetric endomorphismS : V → V such that

η1 = 3λφ, η7 = 3ι(u)ψ, η27 = LSφ.

Since L1φ = 3φ and L1ψ = 4ψ, it follows from equation (8.21) that

dΘ(φ)η1 = λdΘ(φ)L1φ = λL1ψ = 4λψ = 43 ∗φ (3λφ) = 4

3 ∗φ η1. (8.22)

Now define Au ∈ End(V ) by Auv := u× v for v ∈ V . Then

LAuφ = 3ι(u)ψ = η7, LAuψ = ∗φ(3ι(u)ψ) = ∗φη7

56


by (4.18) and (4.19). Hence, it follows from equation (8.21) that

dΘ(φ)η7 = dΘ(φ)LAuφ = LAuψ = ∗φη7. (8.23)

Moreover it follows from equations (8.15) and (8.21)

dΘ(φ)η27 = dΘ(φ)LSφ = LSψ = − ∗φ LSφ = − ∗φ η27. (8.24)

With this understood, equation (8.19) follows from (8.22), (8.23), and (8.24). Thisproves Theorem 8.18.

9. The group Spin(7)

Let W be an 8–dimensional real Hilbert space equipped with a positive triple crossproduct and let Φ ∈ Λ4W ∗ be the Cayley calibration defined by (6.14). We orient W sothat

Φ ∧ Φ > 0

and denote by ∗ : ΛkW ∗ → Λ8−kW ∗ the associated Hodge ∗–operator. Then Φ is self-dual, by Remark 6.21. Recall that, for every unit vector e ∈W , the subspace

Ve := e⊥

is equipped with a cross product

u×e v := u× e× v

and that

Φ = e∗ × φe + ψe, φe := ι(e)Φ ∈ Λ3W ∗, ψe := ∗(e∗ ∧ φe) ∈ Λ4W ∗,

(see Theorem 6.15). The orientation of W is compatible with the decompositionW = 〈e〉 ⊕ Ve (see Remark 6.21).

The group of automorphisms of Φ will be denoted by

G(W,Φ) := g ∈ GL(W ) : g∗Φ = Φ .

By Theorem 7.8, we have G(W,Φ) ⊂ SO(W ) and hence

G(W,Φ) = g ∈ SO(W ) : gu× gv × gw = g(u× v × w) ∀u, v, w ∈W .

For the standard structure Φ0 on R8 in Example 5.32 we denote the structure group bySpin(7) := G(R8,Φ0). By Theorem 7.12, the group G(W,Φ) is isomorphic to Spin(7) forevery positive Cayley-form on an 8–dimensional vector space.

Theorem 9.1. The group G(W,Φ) is a 21–dimensional simple, connected, simply con-nected Lie group. It acts transitively on the unit tangent bundle of the unit sphere and,for every unit vector e ∈ W , the isotropy subgroup Ge := g ∈ G(W,Φ) : ge = e is iso-morphic to G2. Thus there is a fibration

G2 → Spin(7) −→ S7.

57


Proof. The isotropy subgroup Ge is obviously isomorphic to G(Ve, φe) and hence to G2.We prove that G(W,Φ) acts transitively on the unit sphere. Let u, v ∈ W be two unitvectors and choose a unit vector e ∈W which is orthogonal to u and v. By Theorem 8.1,the isotropy subgroup Ge acts transitively on the unit sphere in Ve. Hence, there is anelement g ∈ Ge such that gu = v. That G(W,Φ) acts transitively on the set of pairsof orthonormal vectors now follows immediately from Theorem 8.1. In particular, thereis a fibration G2 → Spin(7) −→ S7. It follows from the homotopy exact sequence ofthis fibration and Theorem 8.1 that Spin(7) is connected and simply connected, and thatπ3(Spin(7)) ∼= Z. Hence, Spin(7) is simple. This proves Theorem 9.1.

Lemma 9.2. Abbreviate

G := G(W,Φ), g := Lie(G) ⊂ so(W ).

The homomorphism ρ : G(W,Φ)→ SO(g⊥) is a nontrivial double cover. Hence, Spin(7)is isomorphic to the universal cover of SO(7).

Proof. Define

I := ξ ∈ g : [ξ, so(W )] ⊂ g .

If ξ ∈ I and η ∈ g, then [[ξ, η], ζ] = −[[η, ζ], ξ] − [[ζ, ξ], η] ∈ g for all ζ ∈ so(W ),and so [ξ, η] ∈ I. Thus I is an ideal in g. Since so(W ) is simple, we have I ( g.Since g is simple, we have I = 0. This implies im ad(ξ) 6⊂ g for 0 6= ξ ∈ g. Sincead(ξ) : so(W ) → so(W ) is skew-adjoint, this implies g⊥ 6⊂ ker ad(ξ) for 0 6= ξ ∈ g. Thismeans that the infinitesimal adjoint action defines an isomorphism g → so(g⊥). Hence,the adjoint action gives rise to a covering map G → SO(g⊥). Since G is connected andsimply connected, this implies that G is the universal cover of SO(g⊥) ∼= SO(7) and thisproves Lemma 9.2.

We examine the action of the group G(W,Φ) on the space

S :=

(u, v, w, x) ∈W∣∣u, v, w, u× v × w, x are orthonormal

.

The space S is a bundle of 3–spheres over a bundle of 5–spheres over a bundle of 6–spheresover a 7–sphere. Hence, it is a compact connected simply connected 21–dimensionalmanifold.

Theorem 9.3. The group G(W,Φ) acts freely and transitively on S .

Proof. Since Spin(7) acts transitively on S7 with isotropy subgroup G2, the result followsimmediately from Theorem 8.2.

Corollary 9.4. The group G(W,Φ) acts transitively on the space of Cayley subspaces ofW .

Proof. This follows directly from Lemma 6.25 and Theorem 9.3.

58


Remark 9.5. For each Cayley subspace H ⊂W choose the orientation such that

volH := Φ|H

is a positive volume form and denote by Λ+H∗ the space of self-dual 2-forms (as inRemark 4.27), by πH : W → H the orthogonal projection, and by

GH := g ∈ G(W,Φ) : gH = H

the isotropy subgroup. Fix a Cayley subspace H ⊂W . Then there is a unique orientationpreserving GH -equivariant isometric isomorphism

TH : Λ+H∗ → Λ+(H⊥)∗.

It is given by

THω := − 12

(∗(Φ ∧ π∗Hω)

)|H⊥ for ω ∈ Λ+H∗ (9.6)

and its inverse is (TH)−1 = TH⊥ . If ω1, ω2, ω3 is a standard basis of Λ+H∗ andτi ∈ Λ+(H⊥)∗ is defined by τi := THωi for i = 1, 2, 3, then the Cayley calibration Φcan be expressed in the form

Φ = π∗HvolH + π∗H⊥volH⊥ −3∑i=1

π∗Hωi ∧ π∗H⊥τi. (9.7)

To see this, choose a standard basis of W as in Example 7.3 such that the vectorse0, e1, e2, e3 form a basis of H, the vectors e4, e5, e6, e7 form a basis of H⊥, and

ω1 = e01 + e23, ω2 = e02 − e13, ω3 = e03 + e12,

τ1 = e45 + e67, τ2 = e46 − e57, τ3 = e47 + e56.

That such a basis exists follows from Theorem 7.12 and Theorem 9.3. It follows alsofrom Theorem 9.3 that a pair (h, h′) ∈ SO(H) × SO(H⊥) belongs to the image of thehomomorphism

GH → SO(H)× SO(H⊥)

if and only if the induced automorphisms of Λ+H∗ and Λ+(H⊥)∗ are conjugate underTH . Hence the map

GH → SO(H)×SO(Λ+H∗) SO(H⊥) : g 7→ [g|H , g|H⊥ ]

is a Lie group isomorphism. Hence, dim GH = 9 and so the Cayley Grassmannian

H := H ⊂W : H is a Cayley subspace ,

which is diffeomorphic to the homogeneous space G(W,Φ)/GH , has dimension 12.

59


Theorem 9.8. There are orthogonal splittings

Λ2W ∗ = Λ27 ⊕ Λ2

21,

Λ3W ∗ = Λ38 ⊕ Λ3

48,

Λ4W ∗ = Λ41 ⊕ Λ4

7 ⊕ Λ427 ⊕ Λ4

35,

where dim Λkd = d and

Λ27 :=

ω ∈ Λ2W ∗ : ∗(Φ ∧ ω) = 3ω

= u∗ ∧ v∗ − ι(u)ι(v)Φ : u, v ∈W ,

Λ221 := ωξ : ξ ∈ g

=ω ∈ Λ2W ∗ : ∗(Φ ∧ ω) = −ω

=ω ∈ Λ2W ∗ : 〈ω, ι(u)ι(v)Φ〉 = ω(u, v) ∀u, v ∈W

,

Λ38 := ι(u)Φ : u ∈W ,

Λ348 :=

ω ∈ Λ3W ∗ : Φ ∧ ω = 0

,

Λ41 := 〈Φ〉,

Λ47 := LξΦ : ξ ∈ so(W ) ,

Λ427 :=

ω ∈ Λ4W ∗ : ∗ω = ω, ω ∧ Φ = 0, ω ∧ LξΦ = 0∀ξ ∈ so(W )

,

Λ435 :=

ω ∈ Λ4W ∗ : ∗ω = −ω

.

Here g := Lie(G(W,Φ)) and, for ξ ∈ so(W ), the 4–form LξΦ ∈ Λ4W ∗ and the 2–form

ωξ ∈ Λ2W ∗ are defined by LξΦ := ddt

∣∣t=0

exp(tξ)∗Φ and ωξ := 〈·, ξ·〉. Each of the spaces

Λkd is an irreducible representation of G(W,Φ).

Proof. By Theorem 9.1, G := G(W,Φ) is simple and so the action of G on g by conjugationis irreducible. Hence, the 21–dimensional subspace Λ2

21 must be contained in an eigenspaceof the operator ω 7→ ∗(Φ ∧ ω) on Λ2W ∗. We prove that the eigenvalue is −1. To see this,we choose a unit vector e ∈W and an element ξ ∈ g with ξe = 0. Let

Ve := e⊥

and denote by ιe : Ve →W and πe : W → Ve the inclusion and orthogonal projection andby ∗e : ΛkV ∗e → Λ7−kV ∗e the Hodge ∗–operator on the subspace. Then

∗(e∗ ∧ π∗eαe) = π∗e ∗e αe ∀ αe ∈ ΛkV ∗e .

Moreover, the alternating forms

φe := ι∗e(ι(e)Φ), ψe := ι∗eΦ

are the associative and coassociative calibrations of Ve. Since ξe = 0, we have ωξ = π∗e ι∗eωξ

and, by Theorem 8.5,

ψe ∧ ι∗eωξ = 0, ∗e(φe ∧ ι∗eωξ) = −ι∗eωξ.

60


Since Φ = e∗ ∧ π∗eφe + π∗eψe, this gives

∗(Φ ∧ ωξ

)= ∗((e∗ ∧ π∗eφe + π∗eψe) ∧ π∗e ι∗eωξ

)= ∗(e∗ ∧ π∗e (φe ∧ ι∗eωξ)

)+ ∗π∗e

(ψe ∧ ι∗eωξ

)= π∗e ∗e (φe ∧ ι∗eωξ) = −π∗e ι∗eωξ = −ωξ.

By Lemma 9.2 the adjoint action of G on g⊥ ⊂ so(W ) is irreducible, and g⊥ is mapped un-der ξ 7→ ωξ onto the orthogonal complement of Λ2

21. Hence, the 7–dimensional orthogonalcomplement of Λ2

21 is also contained in an eigenspace of the operator ω 7→ ∗(Φ∧ω). Sincethis operator is self-adjoint and has trace zero, its eigenvalue on the orthogonal comple-ment of Λ2

21 must be 3 and therefore this orthogonal complement is equal to Λ27. It follows

that the orthogonal projection of ω ∈ Λ2W ∗ onto Λ27 is given by π7(ω) = 1

4 (ω + ∗(Φ ∧ ω)) .Hence, for every nonzero vector e ∈W , we have

Λ27 =

e∗ ∧ u∗ − ι(e)ι(u)Φ : u ∈ e⊥

,

Λ221 =

ω ∈ Λ2W ∗ : 〈ω, ι(e)ι(u)Φ〉 = ω(e, u)∀u ∈ e⊥

.

This proves the decomposition result for Λ2W ∗.

We verify the decomposition of Λ3W ∗. For u ∈W and ω ∈ Λ3W ∗ we have the equation

u∗ ∧ Φ ∧ ω = −ω ∧ ∗ι(u)Φ.

Hence, Φ ∧ ω = 0 if and only if ω is orthogonal to ι(u)Φ for all u ∈W . This shows thatΛ3

48 is the orthogonal complement of Λ38. Since Φ is nondegenerate, we have dim Λ3

8 = 8and, since dim Λ3W ∗ = 56, it follows that dim Λ3

48 = 48.We verify the decomposition of Λ4W ∗. The 4–form g∗Φ is self-dual for every

g ∈ G = G(W,Φ), because Φ is self-dual and G ⊂ SO(W ). This implies that LξΦ isself-dual for every ξ ∈ g = Lie(G). Since SO(W ) has dimension 28 and the isotropysubgroup G of Φ has dimension 21, it follows that the tangent space Λ4

7 to the orbit of Φunder the action of G has dimension 7. As Λ4

1 has dimension 1 and the space of self-dual4–forms has dimension 35, the orthogonal complement of Λ4

1⊕Λ47 in the space of self-dual

4–forms has dimension 27. This proves the dimension and decomposition statements.That the action of G on Λ2

21∼= g is irreducible follows from the fact that G is simple.

Irreducibility of the action on Λ41 is obvious. For Λ3

8∼= W it follows from the fact that G

acts transitively on the unit sphere in W , and for Λ27∼= g⊥ ∼= Λ4

7 it follows from the factthat the isotropy subgroup Ge of a unit vector e ∈W acts transitively on the unit spherein Ve = e⊥. For Λ4

27, Λ435, and Λ3

48 we refer to [Bry87]. This proves Theorem 9.8.

61


Corollary 9.9. For u, v ∈ W denote ωu,v := ι(v)ι(u)Φ = Φ(u, v, ·, ·). Then, for allu, v, x, y ∈W we have

∗ (Φ ∧ u∗ ∧ v∗) = ωu,v, ∗ (Φ ∧ ωu,v) = 3u∗ ∧ v∗ + 2ωu,v, (9.10)

〈ωu,v, ωx,y〉 = 3(〈u, x〉〈v, y〉 − 〈u, y〉〈v, x〉

)+ 2Φ(u, v, x, y), (9.11)

ωu,v ∧ ωx,y ∧ Φ

vol= 6(〈u, x〉〈v, y〉 − 〈u, y〉〈v, x〉

)+ 7Φ(u, v, x, y). (9.12)

Proof. The first equation in (9.10) is a general statement about the Hodge ∗–operator inany dimension. Moreover, by Theorem 9.8, the 2–form u∗ ∧ v∗+ωu,v is an eigenvector ofthe operator ω 7→ ∗(Φ∧ω) with eigenvalue 3. Hence, the second equation in (9.10) followsfrom the first. To prove (9.11), take the inner product of the second equation in (9.10)with x∗ ∧ y∗ and use the identities

〈ωu,v, x∗ ∧ y∗〉 = Φ(u, v, x, y), (9.13)

〈u∗ ∧ v∗, x∗ ∧ y∗〉 = 〈u, x〉〈v, y〉 − 〈u, y〉〈v, x〉, (9.14)

and the fact that the operator ω 7→ ∗(Φ ∧ ω) is self-adjoint. To prove (9.12), we observethat

ωu,v ∧ ωx,y ∧ Φ

vol= 〈ωu,v, ∗(Φ ∧ ωx,y)〉

= 〈ωu,v, 3x∗ ∧ y∗ + 2ωx,y〉= 6(〈u, x〉〈v, y〉 − 〈u, y〉〈v, x〉

)+ 7Φ(u, v, x, y),

where the second equation follows from (9.10) and the last follows from (9.11) and (9.14).This proves Corollary 9.9.

10. Spin structures

This section explains how a cross products in dimension seven, respectively a triplecross products in dimension eight, gives rise to a spin structure and a unit spinor and how,conversely, the cross product or triple cross product can be recovered from these data.We begin the discussion with spin structures and triple cross products in Section 10.1 andthen move on to cross products in Section 10.2.

10.1. Spin structures and triple cross products

Let W be an 8–dimensional oriented real Hilbert space. A spin structure on W is apair of 8–dimensional real Hilbert spaces S± equipped with a vector space homomorphismγ : W → Hom(S+, S−) that satisfies the condition

γ(u)∗γ(u) = |u|21 (10.1)

for all u ∈ W (see [Sal99, Proposition 4.13, Definition 4.32, Example 4.48]). The sign inS± is determined by the condition

γ(e7)∗γ(e6) · · · γ(e1)∗γ(e0) = 1S+ (10.2)

62


for some, and hence every, positively oriented orthonormal basis e0, . . . , e7 ofW (see [Sal99,page 132]). More precisely, consider the 16–dimensional real Hilbert space S := S+⊕S−and define the homomorphism Γ : W → End(S) by

Γ(u) :=

(0 γ(u)

−γ(u)∗ 0

)for u ∈W.

Then equation (10.1) guarantees that Γ extends uniquely to an algebra isomorphismfrom the Clifford algebra C`(W ) to End(S), still denoted by Γ. The complexificationof S gives rise an algebra isomorphism Γc : C`c(W ) → End(Sc) from the complexifiedClifford algebra C`c(W ) := C`(W )⊗R C to the complex endomorphisms of Sc := S⊗R C(see [Sal99, Proposition 4.33]).

Theorem 10.3. Let W be an oriented 8–dimensional real Hilbert space and abbreviateΛk := ΛkW ∗ for k = 0, 1, . . . , 8.

(i) Suppose W is equipped with a positive triple cross product (6.2), let Φ ∈ Λ4 be theCayley calibration defined by (6.14), and assume that Φ ∧ Φ > 0. Define the homo-morphism γ : W → Hom(S+, S−) by

S+ := Λ0 ⊕ Λ27, S− := Λ1 (10.4)

and

γ(u)(λ, ω) := λu∗ + 2ι(u)ω (10.5)

for u ∈ W , λ ∈ R, and ω ∈ Λ27. Then γ is a spin structure on W , i.e., it satis-

fies (10.1) and (10.2). Moreover, the space S+ = Λ0⊕Λ27 of positive spinors contains

a canonical unit vector s = (1, 0) and the triple cross product can be recovered fromthe spin structure and the unit spinor via the formula

γ(u× v × w)s = 〈v, w〉γ(u)s− 〈w, u〉γ(v)s+ 〈u, v〉γ(w)s

− γ(u)γ(v)∗γ(w)s(10.6)

for u, v, w ∈W .(ii) Let γ : W → Hom(S+, S−) be a spin structure and let s ∈ S+ be a unit vector.

Then equation (10.6) defines a positive triple cross product on W and the associatedCayley calibration Φ satisfies Φ ∧ Φ > 0. Since any two spin structures on W areisomorphic, this shows that there is a one-to-one correspondence between positiveunit spinors and positive triple cross products on W that are compatible with theinner product and orientation.

Proof. See page 67.

Assume W is equipped with a positive triple cross product (6.2) and that its Cayleycalibration Φ ∈ Λ4W ∗ in (6.14) satisfies Φ ∧ Φ > 0. Recall that, for every unit vectore ∈W , there is a normed algebra structure on W , defined by (6.19). This normed algebrastructure can be recovered from an intrinsic product map

m : W ×W → Λ0 ⊕ Λ27

63


(which does not depend on e) and an isomorphism γ(e) : Λ0 ⊕ Λ27 → Λ1 (which does

depend on e). The product map is given by

m(u, v) =(〈u, v〉, 1

2 (u∗ ∧ v∗ + ωu,v))

(10.7)

for u, v ∈ W and the isomorphism γ(e) is given by (10.5) with u replaced by e. Hereωu,v := ι(v)ι(u)Φ as in Corollary 9.9.

Lemma 10.8. Let IW : W → W ∗ be the isomorphism induced by the inner product, sothat IW (u) = 〈u, ·〉 = u∗ for u ∈ W . Let γ : W → Hom(S+, S−) and m : W ×W → S+

be defined by (10.5) and (10.7). Then, for all u, v, e ∈W , we have

I−1W (γ(e)m(u, v)) = 〈u, v〉e+ 〈u, e〉v − 〈v, e〉u+ u× e× v, (10.9)

|m(u, v)| = |u||v|, |γ(e)(λ, ω)|2 = |e|2(|λ|2 + |ω|2

). (10.10)

Proof. Equation (10.9) follows directly from the definitions. Moreover, it follows from (9.11)that

|m(u, v)|2 = 〈u, v〉2 + 14 |u∗ ∧ v∗|2 + 1

4 |ωu,v|2

= 〈u, v〉2 + |u ∧ v|2 = |u|2|v|2.

This proves the first equation in (10.10). To prove the second equation in (10.10) weobserve that γ(e)m(e, v) = v and, hence, |γ(e)m(e, v)| = |v| = |m(e, v)| whenever |e| = 1.Since the map W → Λ0 ⊕ Λ2

7 : v 7→ m(e, v) is bijective, this proves Lemma 10.8.

Remark 10.11. If we fix a unit vector e ∈ W and denote v := 2〈e, v〉e − v, then theproduct in (6.19) is given by

uv = −〈u, v〉e+ 〈u, e〉v + 〈v, e〉u+ u× e× v = I−1W (γ(e)m(u, v))

for u, v ∈W .

The next lemma shows that the linear map γ(u) : Λ0 ⊕ Λ27 → Λ1 is dual to the map

m(u, ·) : W → Λ0 ⊕ Λ27 for every u ∈W and that it satisfies equation (10.1).

Lemma 10.12. Let γ : W → Hom(S+, S−) be the homomorphism in (10.4) and (2.21).Then γ satisfies (10.1) and

γ(u)∗v∗ = m(u, v) =(〈u, v〉, 1

2 (u∗ ∧ v∗ + ωu,v))

(10.13)

for all u, v ∈W .

Proof. For u ∈W , λ ∈ R, ω ∈ Λ27, and v ∈W we compute

〈γ(u)(λ, ω), v∗〉 = 〈λu∗ + 2ι(u)ω, v∗〉= λ〈u, v〉+ 2〈ω, u∗ ∧ v∗〉= λ〈u, v〉+ 1

2 〈ω, ωu,v + u∗ ∧ v∗〉 .

The last equation follows from the fact that

π7(u∗ ∧ v∗) = 14 (u∗ ∧ v∗ + ωu,v).

64


This proves (10.13). With this understood, the formula γ(u)∗γ(u) = |u|21 follows directlyfrom (10.10). This proves Lemma 10.12.

Combining the product map m with the triple cross product we obtain an alternatingmulti-linear map τ : W 4 → Λ0 ⊕ Λ2

7 defined by

τ(x, u, v, w) = 14

(m(u× v × w, x)−m(v × w × x, u)

+m(w × x× u, v)−m(x× u× v, w)).

(10.14)

This map corresponds to the four-fold cross product (see Definition 5.33) and has thefollowing properties (see Theorem 5.35).

Lemma 10.15. Let χ : W 4 → Λ27 denote the second component of τ . Then, for all

u, v, w, x ∈W , we have

τ(x, u, v, w) = (Φ(x, u, v, w), χ(x, u, v, w)) ,

Φ(x, u, v, w)2 + |χ(x, u, v, w)|2 = |x ∧ u ∧ v ∧ w|2.

Proof. That the first component of τ is equal to Φ follows directly from the definitions.Moreover, for u, v, w, x ∈W , we have

2χ(x, u, v, w) = (u× v × w)∗ ∧ x∗ + ωu×v×w,x

− (v × w × x)∗ ∧ u∗ − ωv×w×x,u+ (w × x× u)∗ ∧ v∗ + ωw×x×u,v

− (x× u× v)∗ ∧ w∗ − ωx×u×v,w.

(10.16)

We claim that the four rows on the right agree whenever u, v, w, x are pairwise orthogonal.Under this assumption the first two rows remain unchanged if we add to x a multipleof u × v × w. Thus we may assume that x is orthogonal to u, v, w, and u × v × w.By Theorem 9.3, we may therefore assume that W = R8 with the standard triple crossproduct and

u = e0, v = e1, w = e2, x = e4.

In this case a direct computation proves that the first two rows agree. Thus we haveproved that, if u, v, w, x ∈W are pairwise orthogonal, then

τ(x, u, v, w) = m(u× v × w, x).

In this case it follows from (10.10) that

|τ(x, u, v, w)| = |m(u× v × w, x)|= |x||u× v × w|= |x||u||v||w|= |x ∧ u ∧ v ∧ w|.

Since τ is alternating, this proves Lemma 10.15.

65


Lemma 10.17. Let γ : W → Hom(S+, S−) be the homomorphism in (10.4) and (10.5).Then γ satisfies (10.2) and (10.6).

Proof. It follows from (10.1) that 〈γ(u)s, γ(v)s〉 = 〈u, v〉 for all u, v ∈ W . Hence, equa-tion (10.6) is equivalent to

Φ(x, u, v, w) = 〈x, u〉〈v, w〉 − 〈x, v〉〈w, u〉+ 〈x,w〉〈u, v〉− 〈γ(u)∗γ(x)s, γ(v)∗γ(w)s〉

(10.18)

for all x, u, v, w ∈W . Since s = (1, 0) ∈ S+ = Λ0 ⊕ Λ27, we have

γ(u)∗γ(x)s = γ(u)∗x∗ =(〈u, x〉, 1

2 (u∗ ∧ x∗ + ωu,x))

for all u, x ∈W by Lemma 10.12. Hence,

〈γ(u)∗γ(x)s, γ(v)∗γ(w)s〉= 〈u, x〉〈v, w〉+ 1

4 〈u∗ ∧ x∗ + ωu,x, v

∗ ∧ w∗ + ωv,w〉= 〈u, x〉〈v, w〉+ 〈u, v〉〈x,w〉 − 〈u,w〉〈x, v〉+ Φ(u, x, v, w).

Here the last equation follows from Corollary 9.9. This shows that the homomorphism γsatisfies (10.18) and hence also (10.6).

We prove that γ satisfies (10.2). Choose an orthonormal basis e0, . . . , e7 of W in whichΦ has the standard form of Example 7.3. Such a basis exists by Theorem 7.12 becauseΦ is a positive Cayley form, and it is positive because Φ ∧ Φ > 0. Moreover, for anyquadruple of integers 0 ≤ i < j < k < ` ≤ 7, the following are equivalent.

(a) The term ±eijk` appears in the standard basis.(b) Φ(ei, ej , ek, e`) = ±1.(c) ek × ej × ei = ±e`.(d) −γ(ek)γ(ej)

∗γ(ei)s = ±γ(e`)s.

Here the equivalence of (a) and (b) is obvious, the equivalence of (b) and (c) follows fromthe fact that

Φ(ei, ej , ek, e`) = Φ(e`, ek, ej , ei) = 〈ek × ej × ei, e`〉by (7.2), and the equivalence of (c) and (d) follows from equation (10.6). Examining therelevant terms in Example 7.3 we find that

γ(e2)γ(e1)∗γ(e0)s = −γ(e3)s,

hence

γ(e4)γ(e3)∗γ(e2)γ(e1)∗γ(e0)s = −γ(e4)s,

hence

γ(e6)γ(e5)∗γ(e4)γ(e3)∗γ(e2)γ(e1)∗γ(e0)s = −γ(e6)γ(e5)∗γ(e4)s = γ(e7)s,

and hence

γ(e7)∗γ(e6)γ(e5)∗γ(e4)γ(e3)∗γ(e2)γ(e1)∗γ(e0)s = s.

Hence, γ satisfies (10.2) and this proves Lemma 10.17.

66


Proof of Theorem 10.3. Part (i) follows from Lemma 10.12 and Lemma 10.17. To provepart (ii) assume γ : W → Hom(S+, S−) is a spin structure, let s ∈ S+ be a unit vector,and define the multilinear map

W 3 →W : (u, v, w) 7→ u× v × w (10.19)

by (10.6). Then u× v×w = 0 whenever two of the three vectors agree. Hence, it sufficesto verify (6.3) and (6.4) under the assumption that u, v, w are pairwise orthogonal. Inthis case we compute

〈u× v × w, u〉 = 〈γ(u× v × w)s, γ(u)s〉= −〈γ(u)γ(v)∗γ(w)s, γ(u)s〉

= −|u|2〈γ(v)∗γ(w)s, s〉

= −|u|2〈v, w〉 = 0.

and

|u× v × w|2 = |γ(u× v × w)s|2

= |γ(u)γ(v)∗γ(w)s|2

= |u|2|v|2|w|2

= |u ∧ v ∧ w|2.

This shows that the map (10.19) is a triple cross product. To prove that it is positive,choose a quadruple of pairwise orthogonal vectors e, u, v, w ∈ W such that w is alsoorthogonal to e× u× v. Then

γ(e× u× (e× v × w))s = −γ(e)γ(u)∗γ(e× v × w)s

= γ(e)γ(u)∗γ(e)γ(v)∗γ(w)s

= −γ(e)γ(e)∗γ(u)γ(v)∗γ(w)s

= −|e|2γ(u)γ(v)∗γ(w)s

= |e|2γ(u× v × w)s.

Here the first, second, and fifth equalities follow from (10.6) and the third and fourthequalities follow from (10.1). Thus we have proved that the triple cross product (10.19)is positive. That the associated Cayley calibration Φ satisfies Φ ∧ Φ > 0 follows by usinga standard basis and reversing the argument in the proof of Lemma 10.17. This provesTheorem 10.3.

10.2. Spin structures and cross products

Let V be a 7–dimensional oriented real Hilbert space. A spin structure on V isan 8-dimensional real Hilbert space S equipped with a vector space homomorphism

67


γ : V → End(S) that satisfies the conditions

γ(u)∗ + γ(u) = 0, γ(u)∗γ(u) = |u|21 (10.20)

for all u ∈ V (see [Sal99, Definition 4.32]) and

γ(e7)γ(e6) · · · γ(e1) = −1. (10.21)

for some, and hence every, positive orthonormal basis e1, . . . , e7 of V . Equation (10.20)guarantees that the linear map γ : V → End(S) extends uniquely to an algebra homo-morphism γ : C`(V ) → End(S) (see [Sal99, Proposition 4.33]). It follows from (10.21)that the kernel of this extended homomorphism is given by x ∈ C`(V ) : εx = x, whereε := e7 · · · e1 ∈ C`7(V ) for a positive orthonormal basis e1, . . . , e7 of V (see [Sal99, Propo-sition 3.34]). Since ε is an odd element of C`(V ), this implies that the restrictions of γ

to both Cèv(V ) and Còdd(V ) are injective. Since

dim Cèv(V ) = dim Còdd(V ) = dim End(S) = 64,

it follows that γ restricts to an algebra isomorphism from Cèv(V ) to End(S) and to a

vector space isomorphism from Còdd(V ) to End(S).

Theorem 10.22. Let V be an oriented 7–dimensional real Hilbert space.

(i) Suppose V is equipped with a cross product and define the homomorphismγ : V → End(S) by

S := R× V, γ(u)(λ, v) := (−〈u, v〉, λu+ u× v) (10.23)

for λ ∈ R and u, v ∈ V . Then γ is a spin structure on V , i.e., it satisfies (10.20)and (10.21). Moreover, the space S = R × V contains a canonical unit vectors = (1, 0) and the cross product can be recovered from the spin structure and theunit spinor via the formula

γ(u× v)s = γ(u)γ(v)s+ 〈u, v〉s for u, v ∈ V. (10.24)

(ii) Let γ : V → End(S) be a spin structure and let s ∈ S be a unit vector. Then equa-tion (10.24) defines a cross product on V that is compatible with the inner productand orientation. Since any two spin structures on V are isomorphic, this shows thatthere is a one-to-one correspondence between unit spinors and cross products on Vthat are compatible with the inner product and orientation.

Proof. We prove part (i). Thus assume V is equipped with a cross product that iscompatible with the inner product and orientation, and let γ : V → End(S) be givenby (10.23). Then, for u, v, w ∈ V and λ, µ ∈ R, we have

〈(λ, v), γ(u)(µ,w)〉 = µ〈u, v〉 − λ〈u,w〉+ φ(v, u, w).

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This expression is skew-symmetric in (λ, v) and (µ,w) and so γ(u) is skew-adjoint. More-over, for u, v, w ∈ V and µ ∈ R, we have

γ(u)γ(v)(µ,w) + 〈u, v〉(µ,w)

= (−〈u, µv + v × w〉,−〈v, w〉u+ u× (µv + v × w)) + 〈u, v〉(µ,w)

= (−〈u× v, w〉, µ(u× v) + u× (v × w)− 〈v, w〉u+ 〈u, v〉w)

= γ(u× v)(µ,w) + (0,−(u× v)× w − 〈v, w〉u+ 〈u,w〉v)

+ (0,−(v × w)× u− 〈u,w〉v + 〈u, v〉w)

= γ(u× v)(µ,w)− 2(0, [u, v, w]).

Here the last equation follows from (4.1). This proves (10.20) by taking v = −u and (10.24)by taking µ = 1 and w = 0. For the proof of (10.21) it is convenient to use thestandard basis for the standard cross product on V = R7 in Example 2.15. The lefthand side of (10.21) is independent of the choice of the positive orthonormal basis andwe know from general principles that the composition γ(e7) · · · γ(e1) must equal ±1(see [Sal99, Prop 4.34]). The sign can thus be determined by evaluating the composi-tion of the γ(ej) on a single nonzero vector. We leave the verification to the reader. Thisproves part (i).

We prove part (ii). Thus assume that γ : V → End(S) is a spin structure compatiblewith the orientation and let s ∈ S be a unit vector. Then the map

R× V → S : (λ, v) 7→ Ξ(λ, v) := λs+ γ(v)s (10.25)

is an isometric isomorphism, because

|λs+ γ(v)s|2 = |λ|2 + |v|2

by (10.20) and both spaces have the same dimension. For u, v ∈ V the first coordinate ofΞ−1γ(u)γ(v)s is

〈s, γ(u)γ(v)s〉 = −〈u, v〉and so the second coordinate is the vector u × v ∈ V that satisfies (10.24). The mapV × V → V : (u, v) 7→ u × v is obviously bilinear and it is skew symmetric becauseγ(u)γ(v) + γ(v)γ(u) = −2〈u, v〉1 by (10.20). It satisfies (2.3) and (2.10) because

〈u, u× v〉 = 〈γ(u)s, γ(u× v)s〉 = 〈γ(u)s, γ(u)γ(v)s+ 〈u, v〉s〉 = 0,

γ(u× (u× v))s = γ(u)γ(u× v)s = γ(u)(γ(u)γ(v)s+ 〈u, v〉s

)= γ

(〈u, v〉u− |u|2v

)s.

for all u, v ∈ V . Hence, it is a cross product by Lemma 2.9. That it is compatible withthe orientation can be proved by choosing a standard basis as in Example 2.15. Thisproves Theorem 10.22.

We close this section with some useful identities.

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Lemma 10.26. Fix a spin structure γ : V → End(S) that is compatible with the ori-entation and a unit vector s ∈ S, let V × V → V : (u, v) 7→ u × v be the cross productdetermined by (10.24), and let Ξ : R × V → S be the isomorphism in (10.25). Then thefollowing hold:

(i) The spin structure γ is isomorphic to the spin structure in (10.23) via Ξ, i.e., forall λ ∈ R and all u, v ∈ V , we have

Ξ−1γ(u)Ξ(λ, v) = (−〈u, v〉, λu+ u× v) (10.27)

(ii) For all u, v, w ∈ V we have

γ([u, v, w])s+ φ(u, v, w)s+ γ(u)γ(v)γ(w)s

= −〈v, w〉γ(u)s+ 〈w, u〉γ(v)s− 〈u, v〉γ(w)s.(10.28)

(iii) The associative calibration φ ∈ Λ3V ∗ is given by

φ(u, v, w) = −〈s, γ(u)γ(v)γ(w)s〉 (10.29)

and the coassociative calibration ψ = ∗φ ∈ Λ4V ∗ is given by

ψ(u, v, w, x) = −〈s, γ(u)γ(v)γ(w)γ(x)s〉+ 〈v, w〉〈u, x〉 − 〈w, u〉〈v, x〉+ 〈u, v〉〈w, x〉.

(10.30)

Proof. Part (i) follows from (10.24) by direct calculation. By (i) the second displayedformula in the proof of Theorem 10.22 with µ = 0 can be expressed as

γ(u)γ(v)γ(w)s+ 〈u, v〉γ(w)s

= γ(u× v)γ(w)s− 2γ([u, v, w])s

= −2〈u× v, w〉s− 2γ([u, v, w])s− γ(w)γ(u× v)s

= −2φ(u, v, w)s− 2γ([u, v, w])s− γ(w)γ(u)γ(v)s− 〈u, v〉γ(w)s

= −2φ(u, v, w)s− 2γ([u, v, w])s

+ γ(u)γ(w)γ(v)s+ 2〈w, u〉γ(v)s− 〈u, v〉γ(w)s

= −2φ(u, v, w)s− 2γ([u, v, w])s

− γ(u)γ(v)γ(w)s− 2〈v, w〉γ(u)s+ 2〈w, u〉γ(v)s− 〈u, v〉γ(w)s

for all u, v, w ∈ V and this proves (ii). Part (iii) follows from (ii) by taking the innerproduct with s, respectively with γ(x)s (see Lemma 4.8). This proves Lemma 10.26.

11. Octonions and complex linear algebra

Let W be a 2n–dimensional real vector space. An SU(n)–structure on W is a triple(ω, J, θ) consisting of a nondegenerate 2–form ω ∈ Λ2W ∗, an ω–compatible complexstructure J : W →W (so that 〈·, ·〉 := ω(·, J ·) is an inner product), and a complex multi-linear map θ : Wn → C which has norm 2n/2 with respect to the metric determined by

70


ω and J . The archetypal example is W = Cn with the standard symplectic form

ω :=∑j

dxj ∧ dyj ,

the standard complex structure J := i, and the standard (n, 0)–form

θ := dz1 ∧ · · · ∧ dzn.In this section we examine the relation between SU(3)–structures and cross products andbetween SU(4)–structures and triple cross products. We also explain the decompositionsof Theorem 8.5 and Theorem 9.8 in this setting.

Theorem 11.1. Let W be a 6–dimensional real vector space equipped with an SU(3)–structure (ω, J, θ). Then the space V := R ⊕W carries a natural cross product definedby

v × w := (ω(v1, w1), v0Jw1 − w0Jv1 + v1 ×θ w1) (11.2)

for u = (u0, u1), v = (v0, v1) ∈ R⊕W , where v1 ×θ w1 ∈ V is defined by

〈u1, v1 ×θ w1〉 := Re θ(u1, v1, w1)

for all u1 ∈W . The associative calibration of this cross product is

φ := e0 ∧ ω + Re θ ∈ Λ3V ∗ (11.3)

and the coassociative calibration is

ψ := ∗φ = 12ω ∧ ω − e

0 ∧ Im θ ∈ Λ4V ∗. (11.4)

Moreover, the subspaces Λkd ⊂ ΛkV ∗ in Theorem 8.5 are given by

Λ27 = Rω ⊕ e0 ∧ u∗ − ι(u)Im θ : u ∈W,

Λ214 =

τ − e0 ∧ ∗W (τ ∧ Re θ) : τ ∈ Λ2W ∗, τ ∧ ω ∧ ω = 0

,

Λ37 = R · Im θ ⊕

u∗ ∧ ω − e0 ∧ ι(u)Re θ : u ∈W

,

Λ327 = R ·

(3Re θ − 4e0 ∧ ω

)⊕e0 ∧ τ : τ ∈ Λ1,1W ∗, τ ∧ ω ∧ ω = 0

⊕β ∈ Λ2,1W ∗ + Λ1,2W ∗ : β ∧ ω = 0

⊕u∗ ∧ ω + e0 ∧ ι(u)Re θ : u ∈W

.

Proof. For v, w ∈ W we define αv,w ∈ Λ1W ∗ by αv,w := Re θ(·, v, w). Then|αv,w| = |θ(u, v, w)| = |v||w| whenever u, Ju, v, Jv, w, Jw are pairwise orthogonal and|u| = 1. This implies

|αv,w|2 + ω(v, w)2 + 〈v, w〉2 = |v|2|w|2 (11.5)

for all v, w ∈ W . (Add to w a suitable linear combination of v and Jv.) It followsfrom (11.5) by direct computation that the formula (11.2) defines a cross product onR×W . By (11.2) and (11.3), we have φ(u, v, w) = 〈u, v × w〉 so that φ is the associativecalibration of (11.2) as claimed. That φ is compatible with the orientation of R⊕W follows

71


from the fact that ι(e0)φ = ω and ω∧Re θ = 0 so that ι(e0)φ∧ι(e0)φ∧φ = e0∧ω3 = 6vol.The formula (11.4) for ψ := ∗φ follows from the fact that ω ∧ θ = 0 and Im θ = ∗Re θ sothat Re θ ∧ Im θ = 4volW . It remains to examine the subspaces Λkd ⊂ ΛkV ∗ introducedin Theorem 8.5.

The formula for Λ27 follows directly from the formula for φ in (11.3) and the fact that

Λ27 consists of all 2–forms ι(v)φ for v ∈ R⊕W . With v = (1, 0) we obtain ι(v)φ = ω and

with v = (0, Ju) we obtain

ι(v)φ = −e0 ∧ ι(Ju)ω + ι(Ju)Re θ = e0 ∧ u∗ − ι(u)Im θ.

Similarly, the formula for Λ37 follows directly from the formula for ψ in (11.3) and the

fact that Λ37 consists of all 3–forms ι(v)ψ for v ∈ R ⊕W . With v = (−1, 0) we obtain

ι(v)ψ = Im θ and with v = (0,−Ju) we obtain

ι(v)ψ = −(ι(Ju)ω) ∧ ω − e0 ∧ ι(Ju)Im θ = u∗ ∧ ω − e0 ∧ ι(u)Re θ.

To prove the formula for Λ214 we choose α ∈ Λ1W ∗ and τ ∈ Λ2W ∗. Then τ + e0∧α ∈ Λ2

14

if and only if (τ + e0 ∧ α) ∧ ψ = 0. By (11.4), we have

(e0 ∧ α+ τ) ∧ ψ =(e0 ∧ α+ τ

)∧(

12ω ∧ ω − e

0 ∧ Im θ)

= e0 ∧(

12ω ∧ ω ∧ α− τ ∧ Im θ

)+ 1

2τ ∧ ω ∧ ω.

The expression on the right vanishes if and only if τ ∧ω∧ω = 0 and ω∧ω∧α = 2Im θ∧τ .Since α J = 1

2 ∗W (ω ∧ ω ∧ α), the last equation is equivalent to

α = − (∗W (Im θ ∧ τ)) J = − ∗W (Re θ ∧ τ).

To prove the formula for Λ327 we choose τ ∈ Λ2W ∗ and β ∈ Λ3W ∗. Then(

β + e0 ∧ τ)∧ φ = e0 ∧ (τ ∧ Re θ − β ∧ ω) + β ∧ Re θ,(

β + e0 ∧ τ)∧ ψ = e0 ∧

(12τ ∧ ω ∧ ω + β ∧ Im θ

).

Both terms vanish simultaneously if and only if

τ ∧ Re θ = β ∧ ω, β ∧ Re θ = 0, β ∧ Im θ = −1

2τ ∧ ω ∧ ω.

These equations hold in the following four cases.

(a) β = 3λRe θ and τ = −4λω with λ ∈ R.(b) β = 0 and τ ∈ Λ1,1W ∗ with τ ∧ ω ∧ ω = 0.(c) β ∈ Λ1,2W ∗ + Λ2,1W ∗ with β ∧ ω = 0 and τ = 0.(d) β = u∗ ∧ ω and τ = ι(u)Re θ with u ∈W .

In case (d) this follows from (ι(u)Re θ)∧Re θ = 2 ∗ (Ju)∗ = u∗ ∧ω∧ω. The subspacesdetermined by these conditions are pairwise orthogonal and have dimensions 1 in case (a),8 in case (b), 12 in case (c), and 6 in case (d). Thus, for dimensional reasons, their directsum is the space Λ3

27. This proves Theorem 11.1.

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Theorem 11.6. Let W be an 8–dimensional real vector space equipped with an SU(4)–structure (Ω, J,Θ). Then the alternating multi-linear map

Φ := 12Ω ∧ Ω + Re Θ ∈ Λ4W ∗

is a positive Cayley calibration, compatible with the complex orientation and the innerproduct. Moreover, in the notation of Theorem 9.8, we have

Λ27 = RΩ⊕

τ ∈ Λ2,0 + Λ0,2 : ∗ (Re Θ ∧ τ) = 2τ

,

Λ221 =

τ ∈ Λ1,1 : τ ∧ Ω3 = 0

⊕τ ∈ Λ2,0 + Λ0,2 : ∗ (Re Θ ∧ τ) = −2τ

.

Proof. We prove that Φ is compatible with the inner product 〈·, ·〉 := Ω(·, J ·) and thecomplex orientation on W . The associated volume form is 1

24Ω4. Hence, by Lemma 7.4,we must show that

ωu,v ∧ ωu,v ∧ Φ = 14 |u ∧ v|

2Ω4 (11.7)

for all u, v ∈W , where

ωu,v := ι(v)ι(u)Φ = Ω(u, v)Ω− ι(u)Ω ∧ ι(v)Ω + ι(v)ι(u)Re Θ.

To see this, we observe that

ι(v)ι(u)Re Θ ∧ ι(u)Ω ∧ ι(v)Ω ∧ Ω2 = (ι(v)ι(u)Re Θ)2 ∧ Re Θ = 0. (11.8)

If v = Ju, then (11.8) follows from the fact that

ι(u)Ω ∧ ι(Ju)Ω

is a (1, 1)–form and ι(Ju)ι(u)Re Θ = 0. If v is orthogonal to u and Ju, then (11.8)follows from the explicit formulas in Remark 11.9 below. The general case follows fromthe special cases by adding to v a linear combination of u and Ju. Using (11.8) and theidentity

ι(u)Ω ∧ ι(v)Ω ∧ Ω3 = 14Ω(u, v)Ω4

we obtain

ωu,v ∧ ωu,v ∧ Φ = 12Ω(u, v)2Ω4 + 1

2 ι(v)ι(u)Re Θ ∧ ι(v)ι(u)Re Θ ∧ Ω2

− Ω(u, v)ι(u)Ω ∧ ι(v)Ω ∧ Ω3

− 2ι(v)ι(u)Re Θ ∧ ι(u)Ω ∧ ι(v)Ω ∧ Re Θ

= 14Ω(u, v)2Ω4 + 1

2 ι(v)ι(u)Re Θ ∧ ι(v)ι(u)Re Θ ∧ Ω2

− 2ι(v)ι(u)Re Θ ∧ ι(u)Ω ∧ ι(v)Ω ∧ Re Θ.

One can now verify equation (11.7) by first considering the case v = Ju and usingι(Ju)ι(u)Re Θ = 0 (here the last two terms on the right vanish). Next one can ver-ify (11.7) in the case where v is orthogonal to u and Ju by using the SU(4)–symmetryand the explicit formulas in Remark 11.9 below (here the first term on the right van-ishes). Finally, one can reduce the general case to the special cases by adding to v alinear combination of u and Ju.

73


Now recall from Theorem 9.8 that, for every τ ∈ Λ2W ∗, we have

τ ∈ Λ27 ⇐⇒ ∗(Φ ∧ τ) = 3τ,

τ ∈ Λ221 ⇐⇒ ∗(Φ ∧ τ) = −τ.

Since Re Θ ∧ Ω = 0, we have

∗ (Φ ∧ Ω) = 12 ∗ (Ω ∧ Ω ∧ Ω) = 3Ω

and, hence, RΩ ⊂ Λ27. Moreover, Λ2

21 is the image of the Lie algebra g of G(W,Φ) underthe isomorphism

so(W )→ Λ2W ∗ : ξ 7→ ωξ

given by ωξ(u, v) := 〈u, ξv〉. The image of su(W ) under this inclusion is the subspaceτ ∈ Λ1,1W ∗ : τ ∧ Ω3 = 0

and, since SU(W ) ⊂ G(W,Φ), this space is contained in Λ2

21. By considering the standardstructure on C4 we obtain

∗(Ω ∧ Ω ∧ τ) = 2τ

for τ ∈ Λ2,0 + Λ0,2. Hence,

∗(Φ ∧ τ) = 12 ∗ (Ω ∧ Ω ∧ τ) + ∗(Re Θ ∧ τ) = τ + ∗(Re Θ ∧ τ).

for τ ∈ Λ2,0+Λ0,2. Since the operator τ 7→ ∗(Re Θ∧τ) has eigenvalues ±2 on the subspaceΛ2,0 + Λ0,2 the result follows.

Remark 11.9. If (Ω, J,Θ) is the standard SU(4)–structure on W = C4 with coordinates(x1 + iy1, . . . , x4 + iy4), then

Re Θ = dx1 ∧ dx2 ∧ dx3 ∧ dx4 + dy1 ∧ dy2 ∧ dy3 ∧ dy4

− dx1 ∧ dx2 ∧ dy3 ∧ dy4 − dy1 ∧ dy2 ∧ dx3 ∧ dx4

− dx1 ∧ dy2 ∧ dx3 ∧ dy4 − dy1 ∧ dx2 ∧ dy3 ∧ dx4

− dx1 ∧ dy2 ∧ dy3 ∧ dx4 − dy1 ∧ dx2 ∧ dx3 ∧ dy4

and

12Ω ∧ Ω = dx1 ∧ dy1 ∧ dx2 ∧ dy2 + dx3 ∧ dy3 ∧ dx4 ∧ dy4

+ dx1 ∧ dy1 ∧ dx3 ∧ dy3 + dx2 ∧ dy2 ∧ dx4 ∧ dy4

+ dx1 ∧ dy1 ∧ dx4 ∧ dy4 + dx2 ∧ dy2 ∧ dx3 ∧ dy3.

These forms are self-dual. The first assertion in Theorem 11.6 also follows from the factthat the isomorphism R8 → C4 which sends e0, . . . , e7 to

∂/∂x1, ∂/∂y1, ∂/∂x2, ∂/∂y2, ∂/∂x3,−∂/∂y3,−∂/∂x4, ∂/∂y4

pulls back Φ to the standard form Φ0 in Example 5.32.

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Theorem 11.10. Let V be a 7–dimensional real Hilbert space equipped with a crossproduct and its induced orientation. Let φ ∈ Λ3V ∗ be the associative calibration andψ := ∗V φ ∈ Λ4V ∗ the coassociative calibration. Denote W := R⊕V and define Φ ∈ Λ4W ∗

by

Φ := e0 ∧ φ+ ψ.

Then Φ is a positive Cayley-form on W and, in the notation of Theorem 8.5 and Theo-rem 9.8, we have

Λ27W∗ =

e0 ∧ ∗V (ψ ∧ τ) + 3τ : τ ∈ Λ2

7V∗ ,

Λ221W

∗ =e0 ∧ ∗V (ψ ∧ τ)− τ : τ ∈ Λ2V ∗

,

Λ38W∗ = Rφ⊕

ι(u)ψ − e0 ∧ ι(u)φ : u ∈ V

,

Λ348W

∗ = Λ327V

∗ ⊕e0 ∧ τ : τ ∈ Λ2

14V∗⊕ 3ι(u)ψ + 4e0 ∧ ι(u)φ : u ∈ V

,

Λ47W∗ =

e0 ∧ ι(u)ψ − u∗ ∧ φ : u ∈ V

,

Λ427W

∗ =e0 ∧ β + ∗V β : β ∈ Λ3

27V∗ ,

Λ435W

∗ =e0 ∧ β − ∗V β : β ∈ Λ3V ∗

.

Proof. By Theorem 5.4, W is a normed algebra with product (5.6). Hence, by Theo-rem 5.20, W carries a triple cross product (5.26) and Φ is the associated Cayley cali-bration. By Theorem 7.8, Φ is a Cayley form. By (5.24) the triple cross product on Wsatisfies (6.11) with ε = +1 and so is positive (Definition 6.12). Thus, by Theorem 7.12,Φ is positive.

Recall that, by Theorem 9.8, Λ27W∗ and Λ2

21W∗ are the eigenspaces of the operator

∗W (Φ ∧ ·) with eigenvalues 3 and −1 and, by Theorem 8.5, Λ27V∗ and Λ2

14V∗ are the

eigenspaces of the operator ∗V (φ ∧ ·) with eigenvalues 2 and −1. With α ∈ Λ1V ∗ andτ ∈ Λ2V ∗ we have

∗W(Φ ∧

(e0 ∧ α+ τ

))= ∗W

(e0 ∧

(ψ ∧ α+ φ ∧ τ

)+ ψ ∧ τ

)= e0 ∧ ∗V (ψ ∧ τ) + ∗V (φ ∧ τ) + ∗V (ψ ∧ α)

and, hence,

e0 ∧ α+ τ ∈ Λ27W∗ ⇐⇒

∗V (ψ ∧ τ) = 3α,

∗V (φ ∧ τ) + ∗V (ψ ∧ α) = 3τ.

Since ∗V (ψ ∧ ∗V (ψ ∧ τ)) = τ + ∗V (φ ∧ τ), by equation (4.60) in Lemma 4.37, we deducethat e0∧α+τ ∈ Λ2

7W∗ if and only if ∗V (φ∧τ) = 2τ and 3α = ∗V (ψ∧τ). This proves the

formula for Λ27W∗. Likewise, we have e0 ∧α+ τ ∈ Λ2

21W∗ if and only if α = −∗V (ψ∧ τ).

In this case the second equation ∗V (φ ∧ τ) + ∗V (ψ ∧ α) = −τ is automatically satisfied.The formula for the subspace Λ3

8W∗ follows from the fact that it consists of all 3–forms

of the form ι(u)Φ for u ∈W (see Theorem 9.8). Now let τ ∈ Λ2V ∗ and β ∈ Λ3V ∗. Then

e0 ∧ τ + β ∈ Λ348W

∗ ⇐⇒ 0 = Φ ∧(e0 ∧ τ + β

)= e0 ∧ (φ ∧ β + ψ ∧ τ) + ψ ∧ β

75


(see again Theorem 9.8). Hence,

e0 ∧ τ + β ∈ Λ348W

∗ ⇐⇒

φ ∧ β + ψ ∧ τ = 0,

ψ ∧ β = 0.

These conditions are satisfied in the following three cases.

(a) β = 0 and ψ ∧ τ = 0 (or equivalently τ ∈ Λ214V

∗).(b) τ = 0 and φ ∧ β = 0 and ψ ∧ β = 0 (or equivalently β ∈ Λ3

27V∗).

(c) β = 3ι(u)ψ and τ = 4ι(u)φ with u ∈ V .

In the case (a) this follows from the equations ψ ∧ ι(u)ψ = 0 and

3φ ∧ ι(u)ψ + 4ψ ∧ ι(u)φ = 0 (11.11)

for u ∈ V . This last identity can be verified by direct computation using the standardstructure on V = R7 with

φ0 = e123 − e145 − e167 − e246 + e257 − e347 − e356 and

ψ0 = −e1247 − e1256 + e1346 − e1357 − e2345 − e2367 + e4567,

and u := e1 (see the proof of Lemma 4.8). In this case

ι(u)φ0 = e23 − e45 − e67, ι(u)ψ0 = −e247 − e256 + e346 − e357

and so

ψ0 ∧ ι(u)φ0 = 3e234567, φ0 ∧ ι(u)ψ0 = −4e234567.

This proves (11.11). The subspaces determined by the above conditions are pairwiseorthogonal and have dimensions 14 in case (a), 27 in case (b), and 7 in case (c). Thus,for dimensional reasons, their direct sum is Λ3

48W∗.

Now Λ47W∗ is the tangent space of the SO(W )–orbit of Φ. For u ∈ V define the endo-

morphism Au ∈ so(V ) by Auv := u × v. Then, by Remark 4.16, we haveLAuφ = 3ι(u)ψ and LAuψ = −3u∗ ∧ φ. Hence

e0 ∧ ι(u)ψ − u∗ ∧ φ ∈ Λ47W∗

for all u ∈ V . Since Λ47W∗ has dimension 7, each element of Λ4

7W∗ has this form.

Next we recall that Λ427W

∗ is contained in the subspace of self-dual 4–forms, and everyself-dual 4–form can be written as e0 ∧β+ ∗V β with β ∈ Λ3V ∗. By Theorem 9.8 we have

e0 ∧ β + ∗V β ∈ Λ427W

∗ ⇐⇒

β ∧ ∗V φ+ ∗V β ∧ ∗V ψ = 0,

β ∧ ∗V (ι(u)ψ) = ∗V β ∧ ∗V (u∗ ∧ φ) ∀u,⇐⇒ ψ ∧ β = 0, φ ∧ β = 0

⇐⇒ β ∈ Λ327V

∗.

Here the last equivalence follows from Theorem 8.5. This proves the formula for Λ427W

∗.The formula for Λ4

35W∗ follows from the fact that this subspace consists of the anti-self-

dual 4–forms. This proves Theorem 11.10.

76


12. Donaldson–Thomas theory

The motivation for the discussion in these notes came from our attempt to understandRiemannian manifolds with special holonomy in dimensions six, seven, and eight [Bry87,HL82,Joy00] and the basic setting of Donaldson–Thomas theory on such manifolds [DT98,DS11].

12.1. Manifolds with special holonomy

Definition 12.1. Let Y be a smooth 7–manifold and X a smooth 8–manifold. A G2–structure on Y is a nondegenerate 3–form φ ∈ Ω3(Y ); in this case the pair (Y, φ) is calledan almost G2–manifold. An Spin(7)–structure on X is a 4–form Φ ∈ Ω4(X) whichrestricts to a positive Cayley-form on each tangent space; in this case the pair (X,Φ) iscalled an almost Spin(7)–manifold.

Remark 12.2. An almost G2–manifold (Y, φ) admits a unique Riemannian metric and aunique orientation that, on each tangent space, are compatible with the nondegenerate3–form φ as in Definition 3.1 (see Theorem 3.2). Thus each tangent space of Y carries across product

TyY × TyY → TyY : (u, v) 7→ u× vsuch that

φ(u, v, w) = 〈u× v, w〉for all u, v, w ∈ TyY . Moreover, Theorem 8.5 gives rise to a natural splitting of the spaceΩk(Y ) of k–forms on Y for each k.

Remark 12.3. An almost Spin(7)–manifold (X,Φ) admits a unique Riemannian metricthat, on each tangent space, is compatible with the Cayley-form Φ as in Definition 7.1(see Theorem 7.8). Moreover, the positivity hypothesis asserts that the 8–forms

Φ ∧ Φ, ι(v)ι(u)Φ ∧ ι(v)ι(u)Φ ∧ Φ

induce the same orientation whenever u, v ∈ TxX are linearly independent (see Defini-tion 7.11). Thus each tangent space of X carries a positive triple cross product

TxX × TxX × TxX → TxX : (u, v, w) 7→ u× v × wsuch that

Φ(ξ, u, v, w) = 〈ξ, u× v × w〉for all ξ, u, v, w ∈ TxX. Moreover, Theorem 9.8 gives rise to a natural splitting of thespace Ωk(X) of k–forms on X for each k.

Every spin 7–manifold admits aG2–structure [LM89, Theorem 10.6]; concrete examplesare S7 (considered as unit sphere in the octonions), S1×Z where Z is a Calabi–Yau 3–foldand various resolutions of T 7/Γ where Γ is an appropriate finite group, see [Joy00]. A spin

8–manifold X admits a Spin(7)–structure if and only if either χ(/S+

) = 0 or χ(/S−

) = 0[LM89, Theorem 10.7]; concrete examples can be obtained from almost G2–manifolds,Calabi–Yau 4–folds and various resolutions of T 8/Γ.

77


Definition 12.4. An almost G2–manifold (Y, φ) is called a G2–manifold if φ is harmonicwith respect to the Riemannian metric in Remark 12.2 and we say that φ is torsion-free. An almost Spin(7)–manifold (X,Φ) is called a Spin(7)–manifold if Φ is closed(and, hence, harmonic with respect to the Riemannian metric in Remark 12.3) and wesay that Φ is torsion-free.

Remark 12.5. Let (Y, φ) be an almost G2–manifold equipped with the metric of Re-mark 12.2. Then φ is harmonic if and only if φ is parallel with respect to the Levi–Civitaconnection and hence is preserved by parallel transport. It follows that the holonomyof a G2–manifold is contained in the group G2 [FG82]. It also follows that the splittingof Theorem 8.5 is preserved by the Hodge Laplace operator and hence passes on to thede Rham cohomology. Exactly the same holds for an almost Spin(7)–manifold (X,Φ)equipped with the metric of Remark 12.3. The 4–form Φ is closed (and hence harmonic)if and only if it is parallel with respect to the Levi–Civita connection [Bry87]. Thus theholonomy of a Spin(7) manifold is contained in Spin(7) and the splitting of its spaces ofdifferential forms in Theorem 9.8 descends to the de Rham cohomology.

Remark 12.6 (Construction methods). Examples of manifolds with torsion-free G2– orSpin(7)–structures are much harder to construct. There are however a number of construc-tion techniques (all based on gluing methods): Joyce’s generalized Kummer constructionfor G2– and Spin(7)–manifolds [Joy96b, Joy96c, Joy96a, Joy00] based on resolving orb-ifolds of the form T 7/Γ and T 8/Γ; a method of Joyce’s for constructing Spin(7)–manifoldsfrom real singular Calabi–Yau 4–folds [Joy99]; and the twisted connected sum construc-tion invented by Donaldson, pioneered by Kovalev [Kov03], and extended and improvedby Kovalev–Lee [KL11] and Corti–Haskins–Nordstrom–Pacini [CHNP13,CHNP15].

12.2. The gauge theory picture

We close these notes with a brief review of certain partial differential equations arisingin Donaldson–Thomas theory [DT98]. We first discuss the gauge theoretic setting. Let(Y, φ) be a G2–manifold with coassociative calibration ψ := ∗φ and E → Y a G–bundlewith compact semi-simple structure group G. In [DT98] Donaldson and Thomas introducea G2–Chern–Simons functional

CSψ : A (E)→ R

on the space of connections on E. The functional depends on the choice of a referenceconnection A0 ∈ A (E) satisfying FA0 ∧ ψ = 0 and is given by

CSψ(A0 + a) :=1

2

ˆY

(〈dA0a ∧ a〉+

1

3〈a ∧ [a ∧ a]〉

)∧ ψ (12.7)

for a ∈ Ω1(Y,End(E)). The differential of CS has the form

δCSψ(A)a =

ˆN

〈FA ∧ a〉 ∧ ψ

78


for A ∈ A (E) and a ∈ TAA (E) = Ω1(Y,End(E)). Thus a connection A is a critical

point of CSψ if and only if

FA ∧ ψ = 0. (12.8)

By Theorem 8.5 this is equivalent to the equation ∗(FA ∧ φ) = −FA and hence toπ7(FA) = 0. A connection A that satisfies equation (12.8) is called a G2–instanton.As in the case of flat connections on 3–manifolds equation (12.8) becomes elliptic withindex zero after augmenting by a suitable gauge fixing condition (which we do not elab-orate on here). The negative gradient flow lines of the G2–Chern–Simons functional arethe 1–parameter families of connections R → A (E) : t 7→ A(t) satisfying the partialdifferential equation

∂tA = − ∗ (FA ∧ ψ), (12.9)

where FA = FA(t) is understood as the curvature of the connection A(t) ∈ A (E) for afixed value of t. For the study of the solutions of (12.9) it is interesting to observe that,by equation (4.60) in Lemma 4.37, every connection A on Y satisfies the energy identityˆ

Y

|FA|2volY =

ˆY

|FA ∧ ψ|2volY −ˆY

〈FA ∧ FA〉 ∧ φ.

A smooth solution of (12.9) can also be thought of as connection A on the pullbackbundle E of E over R× Y . The curvature of this connection is given by

FA = FA + dt ∧ ∂tA = FA − dt ∧ ∗(FA ∧ ψ).

Hence, it follows from Theorem 9.8 and Theorem 11.10 that FA satisfies

∗ (FA ∧ Φ) = −FA (12.10)

or, equivalently, π7(FA) = 0. Conversely, a connection on E satisfying equation (12.10)can be transformed into temporal gauge and hence corresponds to a solution of (12.9).It is interesting to observe that equation (12.10) makes sense over any Spin(7)–manifold.Solutions of (12.10) are called Spin(7)–instantons. This discussion is completely analo-gous to Floer–Donaldson theory in 3 + 1 dimensions. The hope is that one can constructan analogous quantum field theory in dimension 7 + 1. Moreover, as is apparent fromTheorem 11.1 and Theorem 11.6, this theory will interact with theories in complex di-mensions 3 and 4. The ideas for the real and complex versions of this theory are outlinedin [DT98,DS11].

Remark 12.11. For construction methods and concrete examples of G2–instantons andSpin(7)–instantons we refer to [Wal13,SEW15,Wal15] and [Tan12,Wal16].

12.3. The submanifold picture

There is an analogue of the G2–Chern–Simons functional on the space of 3–dimensionalsubmanifolds of Y , whose critical points are the associative submanifolds of Y and whosegradient flow lines are Cayley submanifolds of R × Y [DT98]. This is the submanifoldpart of the conjectural Donaldson–Thomas field theory.

79


More precisely, let (Y, φ) be a G2–manifold with coassociative calibration ψ = ∗φand let S be a compact oriented 3–manifold without boundary. Denote by F the spaceof smooth embeddings f : S → Y such that f∗φ vanishes nowhere. Then the groupG := Diff+(S) of orientation preserving diffeomorphism of S acts on F by composition.The quotient space

S := F/G

can be identified with the space of oriented 3–dimensional submanifolds of Y that arediffeomorphic to S and have the property that the restriction of φ to each tangent spaceis nonzero; the identification sends the equivalence class [f ] of an element f ∈ F to itsimage f(S).

Given f ∈ F the tangent space of S at [f ] can be identified with the quotient

T[f ]S =Ω0(S, f∗TY )

df ξ : ξ ∈ Vect(S).

If g ∈ G is an orientation preserving diffeomorphism of S, then g∗f := f g is anotherrepresentative of the equivalence class [f ] and the two quotient spaces can be naturally

identified via [f ] 7→ [f g].

Let us fix an element f0 ∈ F and denote by F the universal cover of F based at f0.

Thus the elements of F are equivalence classes of smooth maps f : [0, 1] × S → Y such

that f(0, ·) = f0 and f(t, ·) =: ft ∈ F for all t. Thus we can think of f = ft0≤t≤1

as a smooth path in F starting at f0, and two such paths are equivalent if and only if

they are smoothly homotopic with fixed endpoints. F → F sends f to f := f(1, ·). Theuniversal cover of S is the quotient

S := F/G

where G denotes the group of smooth isotopies [0, 1] → Diff(S) : t 7→ gt starting at the

identity. Now the space F carries a natural G –invariant action functional A : F → Rdefined by

A (f) := −ˆ

[0,1]×Sf∗ψ = −

ˆ 1

0

ˆS

f∗t (ι(∂tft)ψ) dt.

This functional is well defined because ψ is closed and it evidently descends to S . Itsdifferential is the 1–form δA on F given by

δA (f)f = −ˆS

f∗(ι(f)ψ

)This 1–form is G –invariant in that δA (g∗f)g∗f = δA (f)f and horizontal in thatδA (f)dfξ = 0 for ξ ∈ Vect(S). Hence, δA descends to a 1–form on S .

Lemma 12.12. An element [f ] = [ft] ∈ F is a critical point of A if and only if theimage of f := f1 : S → Y is an associative submanifold of Y (that is, each tangentspace is an associative subspace).

80


Proof. We have δA (f) = 0 if and only if ψ(f(x), df(x)ξ, df(x)η, df(x)ζ) = 0 for all

f ∈ Ω0(S, f∗TY ), all x ∈ S, and all ξ, η, ζ ∈ TxS. This means that ψ(u, v, w, ·) = 0 forall q ∈ f(S) and all u, v, w ∈ Tqf(S). By definition of the coassociative calibration ψ inLemma 4.8 this means that [u, v, w] = 0 for all u, v, w ∈ Tqf(S) where

TqY × TqY × TqY → TqY : (u, v, w) 7→ [u, v, w]

denotes the associator bracket defined by (4.1). By Definition 4.6 this means that Tqf(S) isan associative subspace of TqY for all q ∈ f(S). This proves Lemma 12.12.

The tangent space of F at f carries a natural L2 inner product given by⟨f1, f2

⟩L2

:=

ˆS

〈f1, f2〉 f∗φ (12.13)

for f1, f2 ∈ Ω0(S, f∗TY ). This can be viewed as a G –invariant metric on F .

Lemma 12.14. The gradient of A at an element f ∈ F with respect to the innerproduct (12.13) is given by

grad A (f) =[df ∧ df ∧ df ]

f∗φ∈ Ω0(S, f∗TY ),

where [df ∧ df ∧ df ] ∈ Ω3(S, f∗TY ) denotes the 3–form

TxS × TxS × TxS → Tf(x)Y : (ξ, η, ζ) 7→ [df(x)ξ, df(x)η, df(x)ζ].

Proof. The gradient of A at an element f ∈ F is the vector field grad A (f) along fdefined by ˆ

S

〈grad A (f), f〉f∗φ = −ˆS

f∗(ι(f)ψ

)=

ˆS

〈[df ∧ df ∧ df ], f〉.

Here the last equation follows from the identity

−ψ(f , u, v, w) = ψ(u, v, w, f) = 〈[u, v, w], f〉

(see equation (4.9) in Lemma 4.8). This proves Lemma 12.14.

We emphasize that the gradient of A at f is pointwise orthogonal to the image ofdf . This is of course a consequence of the fact that the 1–form δA on F and the innerproduct on TF are G –invariant. Now a negative gradient flow line of A is a smooth map

R× S → Y : (t, x) 7→ ut(x)

that satisfies the partial differential equation

∂tut(x) +[dut(x)e1, dut(x)e2, dut(x)e3]

φ(dut(x)e1, dut(x)e2, dut(x)e3)= 0 (12.15)

for all (t, x) ∈ R × S and every frame e1, e2, e3 of TxS. Moreover, we require of coursethat ut is an embedding for every t and that u∗tφ vanishes nowhere.

81


Lemma 12.16. Let R× S → Y : (t, x) 7→ ut(x) be a smooth map such that ut ∈ F forevery t. Let ξt ∈ Vect(S) be chosen such that

∂tut(x)− dut(x)ξt(x) ⊥ im dut(x) ∀(t, x) ∈ R× S. (12.17)

Then the set

Σ := (t, ut(x)) : t ∈ R, x ∈ S (12.18)

is a Cayley submanifold of R × Y (that is, each tangent space is a Cayley subspace)with respect to the Cayley calibration Φ := dt ∧ φ+ ψ if and only if

∂tut(x)− dut(x)ξt(x) +[dut(x)e1, dut(x)e2, dut(x)e3]

φ(dut(x)e1, dut(x)e2, dut(x)e3)= 0 (12.19)

for every pair (t, x) ∈ R× S and every frame e1, e2, e3 of TxS.

Proof. Fix a pair (t, x) ∈ R×S and choose a basis e1, e2, e3 of TxS. By Theorem 5.20 (iii)the triple cross product of the three tangent vectors

(0, dut(x)e1), (0, dut(x)e2), (0, dut(x)e3)

of Σ is the pair(φ(dut(x)e1, dut(x)e2, dut(x)e3),−[dut(x)e1, dut(x)e2, dut(x)e3]

).

Since this pair is orthogonal to the three vectors (0, dut(x)ei) and its first component isnonzero, it follows that our pair is tangent to Σ if and only if it is a scalar multiple of thepair(1, ∂tut(x) − dut(x)ξt(x)). This is the case if and only if (12.19) holds. Hence, it fol-lows from Lemma 6.25 that Σ is a Cayley submanifold of R× Y if and only if u satisfiesequation (12.19). This proves Lemma 12.16.

Lemma 12.16 shows that every negative gradient flow line of A determines a Cayleysubmanifold Σ ⊂ R×Y via (12.18) and, conversely, every Cayley submanifold Σ ⊂ R×Y ,with the property that the projection Σ→ R is a proper submersion, can be parametrizedas a negative gradient flow line of A (for some S). Thus the negative gradient trajectoriesof A are solutions of an elliptic equation, after taking account of the action of the infinitedimensional reparametrization group G . They minimize the energy

E(u, ξ) := 12

ˆ ∞−∞

ˆS

(|∂tut − dutξt|2 +

∣∣∣∣ [dut ∧ dut ∧ dut]u∗tφ

∣∣∣∣2)u∗tφdt

= 12

ˆ ∞−∞

ˆS

∣∣∣∣∂tut − dutξt +[dut ∧ dut ∧ dut]

u∗tφ

∣∣∣∣2u∗tφdt+

ˆR×S

u∗ψ.

For studying the solutions of (12.19) it will be interesting to introduce the energy densityef : S → R of an embedding f ∈ F via

ef (x) :=det(〈df(x)ei, df(x)ej〉i,j=1,2,3

)φ(df(x)e1, df(x)e2, df(x)e3)2

82


for every x ∈ S and every frame e1, e2, e3 of TxS. Then efg = ef g for every (orientationpreserving) diffeomorphism g of S and so the energy

E (f) :=

ˆS

eff∗φ (12.20)

is a G –invariant function on F . Moreover, it follows from Lemma 4.4 that

E (f) =

ˆS

∣∣∣∣ [df ∧ df ∧ df ]

f∗φ

∣∣∣∣2f∗φ+

ˆS

f∗φ.

If φ is closed, then the last term on the right is a topological invariant. Moreover, thefirst term vanishes if and only if f is a critical point of the action functional A . Thus thecritical points of A are also the absolute minima of the energy E (in a given homologyclass).

12.4. Outlook: difficulties and new phenomena

These observations are the starting point of a conjectural Floer–Donaldson type theoryin dimensions seven and eight, as outlined in the paper by Donaldson and Thomas [DT98].The analytical difficulties one encounters when making this precise are formidable, in-cluding non-compactness phenomena in codimension four [Tia00] and two in the gaugetheory and submanifold theory respectively. The work of Donaldson and Segal [DS11]explains that this leads to new geometric phenomena linking the gauge theory and thesubmanifold theory. It is now understood that neither the naive approach to count-ing G2–instantons [DS11, Wal12] nor that of counting associative submanifolds [Nor13]can work on their own. There are, however, ideas of how the theories outlined in Sec-tion 12.2 and Section 12.3 have to be combined and extended to obtain new invari-ants [DS11,HW15].

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ETH Zurich, Department of Mathematics, 8092 Zrich, Switzerland

E-mail address: [email protected]

Massachusetts Institute of Technology, Department of Mathematics, 77 Massachusetts

Avenue, Cambridge MA 02139-4307E-mail address: [email protected]

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